Algebra 1

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. List the number sets, from smallest to largest, that contain the natural numbers.
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   Natural, whole, integer, rational, real

2. Classify \( \sqrt{81} \) into every set it belongs to.
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   \( \sqrt{81} = 9 \), a counting number. Natural, whole, integer, rational, real

3. Classify \( -5 \) into every set it belongs to.
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   Negative, so not whole or natural. It is an integer, and \( -5 = \frac{-5}{1} \). Integer, rational, real

4. Name the property shown by \( 3(x + 5) = 3x + 15 \).
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   The distributive property

5. Name the property shown by \( 7 \cdot 4 = 4 \cdot 7 \).
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   The commutative property of multiplication

6. Is \( \sqrt{16} + \sqrt{2} \) rational or irrational? Justify your answer.
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   Simplify first: \( \sqrt{16} = 4 \), which is rational, while \( \sqrt{2} \) is irrational. A rational plus an irrational is always irrational, because if the sum were rational you could subtract the rational part and be left with a rational value for \( \sqrt{2} \), which is impossible. Irrational

7. Give a counterexample showing that subtraction is not associative.
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   Test \( (10 - 4) - 3 \) against \( 10 - (4 - 3) \). The left gives \( 6 - 3 = 3 \). The right gives \( 10 - 1 = 9 \). Since \( 3 \neq 9 \), regrouping changed the result, so subtraction is not associative. One counterexample is enough to disprove a universal claim. \( (10 - 4) - 3 = 3 \) but \( 10 - (4 - 3) = 9 \)

8. A student says every square root is irrational. Give two counterexamples and state the rule correctly.
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   \( \sqrt{25} = 5 \) and \( \sqrt{100} = 10 \), both rational. The correct rule is that the square root of a whole number is rational exactly when that number is a perfect square, and irrational otherwise. There is nothing in between, which is why simplifying before classifying matters. \( \sqrt{25} \) and \( \sqrt{100} \); a square root is rational when the radicand is a perfect square

9. Name the property justifying each step: \( 2(x + 3) + 4x = 2x + 6 + 4x = 2x + 4x + 6 = (2 + 4)x + 6 = 6x + 6 \).
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   The first step distributes the 2 across the sum, which is the distributive property. The second moves the 6 past \( 4x \), changing the order of terms, which is the commutative property of addition. The third groups the two \( x \) terms and factors out the \( x \), which is the distributive property again, read right to left. The last is arithmetic. Distributive, commutative (addition), distributive, then arithmetic

10. Explain why 0 has no multiplicative inverse, and why every other real number does.
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    The multiplicative inverse of \( a \) is the number \( \frac{1}{a} \) satisfying \( a \cdot \frac{1}{a} = 1 \). For any nonzero \( a \) that number exists and is a real number. For \( a = 0 \), the requirement would be a number \( b \) with \( 0 \cdot b = 1 \). But zero times anything is zero, so no such \( b \) exists, and the requirement is not merely hard to satisfy but impossible. This is exactly why division by zero is undefined: dividing by \( a \) means multiplying by its multiplicative inverse, and zero has none. No number multiplied by 0 gives 1, since zero times anything is zero, which is why division by zero is undefined

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( 3 + 4 \cdot 5 \).
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   Multiplication first: \( 4 \cdot 5 = 20 \), then \( 3 + 20 \). 23

2. Evaluate \( (3 + 4) \cdot 5 \).
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   Grouping first: \( 3 + 4 = 7 \), then \( 7 \cdot 5 \). 35

3. Evaluate \( -5^2 \) and \( (-5)^2 \).
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   The exponent binds tighter than the negative sign, so the first is \( -(25) \). In the second the brackets make the base \( -5 \). \( -25 \) and \( 25 \)

4. Evaluate \( 2x^2 - 3x \) when \( x = -2 \).
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   \( 2(-2)^2 - 3(-2) = 2(4) + 6 = 8 + 6 \). 14

5. Evaluate \( \dfrac{a + b}{a - b} \) when \( a = 9 \) and \( b = 3 \).
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   Top: \( 9 + 3 = 12 \). Bottom: \( 9 - 3 = 6 \). Then \( 12 \div 6 \). 2

6. Evaluate \( 24 \div 4 \cdot 3 \) and explain why the answer is not 2.
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   Multiplication and division have equal priority and are done left to right, so \( 24 \div 4 = 6 \) first, then \( 6 \cdot 3 = 18 \). Getting 2 comes from doing the multiplication first, treating \( 4 \cdot 3 = 12 \) as though multiplication outranked division. It does not; they are one level worked left to right. 18; division and multiplication have equal priority and go left to right

7. Evaluate \( -x^2 + 4x - 1 \) when \( x = -3 \).
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   Substitute with brackets: \( -(-3)^2 + 4(-3) - 1 \). The exponent applies to \( -3 \) inside the bracket, giving 9, and the leading minus then negates it: \( -9 \). Next \( 4(-3) = -12 \). So \( -9 - 12 - 1 \). \( -22 \)

8. Evaluate \( \dfrac{2(x + 3)^2}{x - 1} \) when \( x = 5 \).
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   Numerator: \( (5 + 3) = 8 \), then \( 8^2 = 64 \), then \( 2(64) = 128 \). Denominator: \( 5 - 1 = 4 \). Then \( 128 \div 4 \). 32

9. A student evaluates \( 5 - 2(x - 4) \) at \( x = 1 \) and gets 9. Find their error and the correct value.
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   Correctly: \( 5 - 2(1 - 4) = 5 - 2(-3) = 5 + 6 = 11 \). Getting 9 comes from subtracting the 2 from the 5 first, computing \( 3(x-4) = 3(-3) = -9 \), and then taking the sign wrong; either way the error is treating the subtraction as higher priority than the multiplication. Multiplication comes first, and the \( -2 \) travels with the bracket. 11; they subtracted before multiplying

10. Explain why the order of operations is a convention rather than something that could be proved, and what would go wrong without one.
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    Nothing in arithmetic forces \( 3 + 4 \cdot 5 \) to mean \( 3 + (4 \cdot 5) \) rather than \( (3 + 4) \cdot 5 \). The symbols alone are ambiguous, and mathematicians chose one reading so that written expressions have a single meaning. The choice is not arbitrary in spirit, since giving multiplication priority lets a polynomial like \( 3x^2 + 2x + 1 \) be written without brackets, which is why the convention is convenient. Without an agreed order the same expression would evaluate differently for different readers, so no formula could be communicated reliably, and every expression would need brackets everywhere. The symbols are ambiguous on their own; the convention makes written expressions unambiguous, and without it the same expression would mean different things to different readers

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Convert 5 feet to inches.
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   \( 5 \text{ ft} \times \frac{12 \text{ in}}{1 \text{ ft}} \), feet cancels. 60 inches

2. Convert 2.5 hours to minutes.
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   \( 2.5 \times 60 \). 150 minutes

3. A recipe uses 3 cups per batch. How many cups for 7 batches?
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   \( 7 \text{ batches} \times \frac{3 \text{ cups}}{1 \text{ batch}} \). 21 cups

4. State the unit of the rate "distance divided by time".
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   A unit of distance per unit of time, such as miles per hour or meters per second

5. Convert 120 minutes to seconds.
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   \( 120 \times 60 \). 7,200 seconds

6. A printer produces 22 pages per minute. How many pages in 45 minutes, and how long to print 1,000 pages?
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   For the first: \( 45 \text{ min} \times \frac{22 \text{ pages}}{1 \text{ min}} = 990 \) pages. For the second, flip the rate so pages cancels: \( 1000 \text{ pages} \times \frac{1 \text{ min}}{22 \text{ pages}} = 45.45\ldots \) minutes, about 45.5 minutes. 990 pages; about 45.5 minutes

7. Convert 30 miles per hour to feet per second.
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   \[ \frac{30 \text{ mi}}{1 \text{ h}} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \] Miles cancels and hours cancels, leaving feet per second. \( \frac{30 \times 5280}{3600} = \frac{158{,}400}{3600} = 44 \). 44 feet per second

8. A tank fills at 12 gallons per minute. Express this in gallons per hour and explain why the number gets larger.
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   \( \frac{12 \text{ gal}}{1 \text{ min}} \times \frac{60 \text{ min}}{1 \text{ h}} = 720 \) gallons per hour. The number grows because an hour is a much longer period than a minute, so the same rate delivers 60 times as much in it. The quantity of water per unit time has not changed; only the unit it is reported in has. 720 gallons per hour; the time unit is 60 times larger

9. A student converts 3 hours to seconds and gets 180. Diagnose the error.
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   They converted hours to minutes and stopped. \( 3 \text{ h} \times \frac{60 \text{ min}}{1 \text{ h}} = 180 \) minutes, and a second conversion is still needed: \( 180 \text{ min} \times \frac{60 \text{ s}}{1 \text{ min}} = 10{,}800 \) seconds. Writing the units on the page would have caught it immediately, because the units after the first step read "minutes" rather than "seconds". 10,800 seconds; they stopped after converting to minutes

10. A car uses fuel at 28 miles per gallon and fuel costs $3.60 per gallon. Find the cost per mile, and explain how the units confirm the setup.
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    The target unit is dollars per mile, so dollars must end on top and miles on the bottom. Start with the price, which already has dollars on top, and multiply by the reciprocal of the fuel economy so that gallons cancels and miles lands on the bottom: \[ \frac{3.60 \text{ dollars}}{1 \text{ gal}} \times \frac{1 \text{ gal}}{28 \text{ mi}} = \frac{3.60}{28} \frac{\text{dollars}}{\text{mi}} \] Gallons cancels, leaving dollars per mile as required. \( 3.60 \div 28 = 0.12857\ldots \), so about $0.129 per mile, or roughly 12.9 cents. Note that multiplying the two figures instead would have given units of dollars times miles per gallon squared, which is meaningless, and that is the signal that the setup was wrong. About $0.129 per mile; gallons cancels to leave dollars per mile

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write an expression for "seven more than a number \( n \)".
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   \( n + 7 \)

2. Write an expression for "four less than a number \( n \)".
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   "Less than" reverses the order. \( n - 4 \)

3. Write an expression for "the product of 5 and a number \( x \)".
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   \( 5x \)

4. Write an expression for "twice the sum of \( x \) and 8".
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   "Sum" marks a grouped quantity. \( 2(x + 8) \)

5. Write an expression for "30 percent of a number \( p \)".
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   \( 0.30p \)

6. Write expressions for "three less than twice a number" and "twice, three less than a number", and explain why they differ.
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   The first is \( 2n - 3 \): double the number, then take 3 away. The second is \( 2(n - 3) \): take 3 away first, then double, so the 3 is doubled as well. Expanding the second gives \( 2n - 6 \), which is 3 smaller than the first for every value of \( n \). The phrase order tells you which operation happens inside the bracket. \( 2n - 3 \) and \( 2(n - 3) = 2n - 6 \)

7. A plumber charges a $65 callout fee plus $48 per hour. Write an expression for the cost of a job lasting \( h \) hours, and state what each part represents.
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   The callout fee is charged once regardless of duration, so it is a constant. The hourly charge accumulates, so it is multiplied by the number of hours. \( C = 48h + 65 \), where \( 48h \) is the labor cost in dollars and 65 is the fixed callout fee in dollars. \( C = 48h + 65 \)

8. A rectangle's length is 4 cm more than three times its width. Write expressions for the length and for the perimeter in terms of the width \( w \).
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   "Three times its width" is \( 3w \), and "4 more than" that is \( 3w + 4 \), so the length is \( 3w + 4 \) cm. Perimeter is twice the length plus twice the width: \[ P = 2(3w + 4) + 2w = 6w + 8 + 2w = 8w + 8 \] Length \( 3w + 4 \) cm; perimeter \( 8w + 8 \) cm

9. A student writes "8 less than a number" as \( 8 - n \). Show with a specific value why this is wrong.
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   Take the number to be 20. Eight less than 20 is 12. The student's expression gives \( 8 - 20 = -12 \), which is not 12. The correct expression \( n - 8 \) gives \( 20 - 8 = 12 \). The phrase names the amount removed and the thing it is removed from, in that order, so the subtraction is written in the opposite order to the words. \( n - 8 \); at \( n = 20 \) the student's version gives \( -12 \) instead of 12

10. A phone plan charges $25 per month plus $0.10 for each text over 200. Write an expression for the monthly cost when \( t \) texts are sent, assuming \( t \gt 200 \), and explain why the expression is not simply \( 25 + 0.10t \).
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    Only the texts beyond 200 are charged, so the billable count is \( t - 200 \), not \( t \). The cost is therefore \[ C = 25 + 0.10(t - 200) = 25 + 0.10t - 20 = 0.10t + 5 \] Writing \( 25 + 0.10t \) would charge for all texts including the 200 already covered by the monthly fee, overcharging by \( 0.10 \times 200 = \$20 \) every month. Checking at \( t = 200 \) confirms the correct version: it gives \( 0.10(200) + 5 = \$25 \), exactly the base fee, as it should. Note the expression is only valid for \( t \gt 200 \); below that the cost is a flat $25. \( C = 25 + 0.10(t - 200) = 0.10t + 5 \) for \( t \gt 200 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Expand \( 4(x + 6) \).
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   \( 4x + 24 \)

2. Expand \( -3(x + 5) \).
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   Both terms are multiplied by \( -3 \). \( -3x - 15 \)

3. Expand \( -2(4x - 7) \).
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   \( -2 \times 4x = -8x \) and \( -2 \times (-7) = +14 \). \( -8x + 14 \)

4. Simplify \( 9y - 4y + 2y \).
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   All like terms; add the coefficients: \( 9 - 4 + 2 = 7 \). \( 7y \)

5. Simplify \( -(x - 9) \).
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   The bare minus is \( -1 \) times the bracket. \( -x + 9 \)

6. Simplify \( 3(2x + 4) - 2(x - 5) \).
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   First bracket: \( 6x + 12 \). Second: \( -2 \times x = -2x \) and \( -2 \times (-5) = +10 \), giving \( -2x + 10 \). Together: \( 6x + 12 - 2x + 10 \). Combining: \( 6x - 2x = 4x \) and \( 12 + 10 = 22 \). \( 4x + 22 \)

7. Simplify \( 5 - 2(3x - 4) + x \).
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   Distribute the \( -2 \): \( -2 \times 3x = -6x \) and \( -2 \times (-4) = +8 \). So the expression is \( 5 - 6x + 8 + x \). Combine \( x \) terms: \( -6x + x = -5x \). Combine constants: \( 5 + 8 = 13 \). \( -5x + 13 \)

8. Simplify \( 4(2a - 3b) - 3(a - 2b) \).
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   First bracket: \( 8a - 12b \). Second: \( -3 \times a = -3a \) and \( -3 \times (-2b) = +6b \), giving \( -3a + 6b \). Together: \( 8a - 12b - 3a + 6b \). The \( a \) terms give \( 8a - 3a = 5a \); the \( b \) terms give \( -12b + 6b = -6b \). These cannot be combined with each other, since \( a \) and \( b \) are different variables. \( 5a - 6b \)

9. A student simplifies \( 7 - 3(x - 2) \) to \( 4x - 8 \). Find both errors and give the correct answer.
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   Two things went wrong. First, they subtracted 3 from 7 before distributing, treating the expression as \( 4(x - 2) \); multiplication must be done before subtraction. Second, they distributed a positive 3 rather than the \( -3 \) that actually sits in front of the bracket. Correctly: \( -3 \times x = -3x \) and \( -3 \times (-2) = +6 \), so the expression is \( 7 - 3x + 6 = -3x + 13 \). \( -3x + 13 \); they subtracted before distributing and dropped the negative sign

10. Explain why \( 3x + 4y \) cannot be simplified to \( 7xy \), using a substitution to make the point.
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    Terms can only be combined when they have identical variable parts, because combining is an application of the distributive property in reverse: \( 3x + 4x = (3 + 4)x \) works precisely because the same \( x \) can be factored out. With \( 3x + 4y \) there is no common variable factor, so nothing can be pulled out and the terms stay separate. A substitution settles it: at \( x = 2 \) and \( y = 5 \), the original is \( 3(2) + 4(5) = 6 + 20 = 26 \), while \( 7xy \) gives \( 7(2)(5) = 70 \). Different values for the same inputs means the two expressions are not equal. They are unlike terms with no common factor; at \( x = 2, y = 5 \) the original is 26 while \( 7xy \) is 70

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. How many terms does \( 5x^2 - 3x + 8 \) have?
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   Three

2. State the coefficient of \( x \) in \( 9 - 6x \).
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   The sign belongs to the coefficient. \( -6 \)

3. List the factors of the term \( 12x^3 \).
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   12 and three factors of \( x \); the numerical factor 12 further factors as \( 2 \cdot 2 \cdot 3 \)

4. Write \( x^2 - 25 \) in factored form.
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   A difference of squares with \( a = x \), \( b = 5 \). \( (x - 5)(x + 5) \)

5. Identify the common factor in \( 6x^2 + 9x \).
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   \( 3x \)

6. Factor \( 49y^2 - 16 \) and verify by expanding.
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   \( 49y^2 = (7y)^2 \) and \( 16 = 4^2 \), so this is \( a^2 - b^2 \) with \( a = 7y \) and \( b = 4 \). It factors as \( (7y - 4)(7y + 4) \). Expanding: \( 49y^2 + 28y - 28y - 16 = 49y^2 - 16 \). The middle terms cancel, confirming it. \( (7y - 4)(7y + 4) \)

7. The expression \( 3(x + 2)^2 - 12 \) has the structure of a difference of squares if you first factor something out. Factor it completely.
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   Both terms share a factor of 3: \( 3\big[(x + 2)^2 - 4\big] \). The bracket is now a difference of squares with \( a = (x + 2) \) and \( b = 2 \), so it factors as \( \big[(x + 2) - 2\big]\big[(x + 2) + 2\big] = (x)(x + 4) \). Altogether \( 3x(x + 4) \). Check by expanding: \( 3x^2 + 12x \), and the original expands to \( 3(x^2 + 4x + 4) - 12 = 3x^2 + 12x + 12 - 12 = 3x^2 + 12x \). They agree. \( 3x(x + 4) \)

8. Explain what different information the forms \( x^2 - 6x + 8 \) and \( (x - 2)(x - 4) \) each make immediately visible.
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   They are the same expression written two ways, and each puts a different feature on the surface. Standard form shows the constant term 8, which is the value of the expression when \( x = 0 \), so it gives the \( y \) intercept of the graph at a glance. Factored form shows that the expression is zero exactly when \( x = 2 \) or \( x = 4 \), so it gives the \( x \) intercepts at a glance. Neither is simpler; you choose the one that answers the question being asked. Standard form shows the \( y \) intercept 8; factored form shows the zeros 2 and 4

9. Explain why \( x^2 + 9 \) cannot be factored over the real numbers, while \( x^2 - 9 \) can.
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   The difference of squares pattern works because the middle terms cancel: \( (x - 3)(x + 3) = x^2 + 3x - 3x - 9 = x^2 - 9 \). For that cancellation to happen, the two factors must have opposite signs, and multiplying them then produces a minus sign in front of \( b^2 \). No pair of real factors gives \( +9 \) with no middle term, because any such pair would have to be \( (x + a)(x + b) \) with \( a + b = 0 \) and \( ab = 9 \), meaning \( b = -a \) and \( -a^2 = 9 \), which has no real solution. A sum of squares is therefore prime over the reals. A difference of squares needs factors of opposite sign, which produces a minus; no real pair gives \( +9 \) with no middle term

10. Factor \( x^4 - 16 \) completely, and explain why stopping at two factors would be incomplete.
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    \( x^4 = (x^2)^2 \) and \( 16 = 4^2 \), so this is a difference of squares: \( x^4 - 16 = (x^2 - 4)(x^2 + 4) \). Stopping here would be incomplete because the first factor is itself a difference of squares: \( x^2 - 4 = (x - 2)(x + 2) \). The second factor, \( x^2 + 4 \), is a sum of squares and does not factor over the reals, so it stays. The complete factorization is \( (x - 2)(x + 2)(x^2 + 4) \). The word "completely" means every factor must be examined again after each step, and the process stops only when no factor can be broken down further. \( (x - 2)(x + 2)(x^2 + 4) \); the first factor of the two-factor form is itself a difference of squares

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Round 3.7482 to two decimal places.
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   The third decimal is 8, so round the second up. 3.75

2. Round 1,472 to the nearest hundred.
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   1,500

3. A price is calculated as $14.2857. State it as an amount of money.
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   $14.29

4. Give the units of an answer found by dividing dollars by hours.
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   Dollars per hour

5. A calculation gives 8.4 people. State a sensible answer.
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   People are counted in whole numbers, and the right direction depends on the context; if these are people who must be seated or served, round up. 8 people, or 9 if everyone must be accommodated

6. A rope 12.5 m long is cut into pieces 1.8 m long. How many whole pieces, and how much is left over?
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   \( 12.5 \div 1.8 = 6.944\ldots \), so 6 whole pieces can be cut; a seventh would require 12.6 m. Used length: \( 6 \times 1.8 = 10.8 \) m. Remaining: \( 12.5 - 10.8 = 1.7 \) m. Here the context forces rounding down, the opposite of the bus problem, because a partial piece is not a usable piece. 6 pieces, with 1.7 m left over

7. A rectangle measures 4.2 cm by 7.9 cm. Find its area and state it to a sensible precision.
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   \( 4.2 \times 7.9 = 33.18 \) cm². Each measurement is given to two significant figures, so the product should not be reported to four. Rounding to two significant figures gives 33 cm². Reporting 33.18 would claim the measurements were accurate to a hundredth of a centimeter, which the given data does not support. About 33 cm²

8. A student computes \( \frac{2}{3} \times 900 \) by first rounding \( \frac{2}{3} \) to 0.67. Find their answer, the correct answer, and the size of the error.
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   Their answer: \( 0.67 \times 900 = 603 \). Correct: \( \frac{2}{3} \times 900 = \frac{1800}{3} = 600 \) exactly. The error is 3, which is small as a fraction but is a whole 3 units in a context where the answer might be a count. The rounding error of 0.00333 in the fraction was multiplied by 900, magnifying it 900 times. This is exactly why rounding belongs at the end. 603 against an exact 600; an error of 3

9. A car travels 247 miles on 9.1 gallons. Compute the fuel economy and justify the precision you report.
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   \( 247 \div 9.1 = 27.142857\ldots \) miles per gallon. The fuel measurement is given to two significant figures, which is the weaker of the two inputs and therefore limits the result. Reporting 27.142857 would be absurd; even 27.14 claims more than the data supports. Rounding to two significant figures gives 27 miles per gallon, and quoting 27.1 would be defensible if the 9.1 is trusted to that tenth. About 27 miles per gallon, limited by the two-figure fuel measurement

10. A theater has 340 seats. Tickets cost $18.50 and the show costs $4,000 to stage. Find the number of tickets that must be sold to break even, and explain why the rounding direction is not the usual one.
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    Break even means revenue equals cost: \( 18.50t = 4000 \), so \( t = 4000 \div 18.50 = 216.216\ldots \) tickets. Ordinary rounding would give 216, but selling 216 tickets brings in \( 216 \times 18.50 = \$3{,}996 \), which is $4 short of covering the cost. The context requires meeting or exceeding the cost, so the answer must round up to 217 tickets, giving \( 217 \times 18.50 = \$4{,}014.50 \). The decimal .216 would normally round down; the situation overrides that, because a shortfall is not acceptable. It is also worth noting 217 is well under the 340 seats available, so break even is achievable. 217 tickets; the context requires covering the full cost, so it rounds up despite the decimal

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( x + 9 = 17 \).
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   Subtract 9 from both sides. \( x = 8 \)

2. Solve \( 5x = 40 \).
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   Divide both sides by 5. \( x = 8 \)

3. Solve \( 2x - 6 = 14 \).
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   Add 6: \( 2x = 20 \). Divide by 2. \( x = 10 \)

4. Solve \( \dfrac{x}{4} = 7 \).
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   Multiply both sides by 4. \( x = 28 \)

5. Solve \( -3x = 21 \).
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   Divide both sides by \( -3 \). \( x = -7 \)

6. Solve \( 4x + 11 = 3 \) and check your answer.
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   Subtract 11: \( 4x = -8 \). Divide by 4: \( x = -2 \). Check: \( 4(-2) + 11 = -8 + 11 = 3 \). Correct. \( x = -2 \)

7. Solve \( \dfrac{x}{5} - 3 = 2 \) and name the property used at each step.
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   Add 3 to both sides, by the addition property of equality: \( \frac{x}{5} = 5 \). Multiply both sides by 5, by the multiplication property of equality: \( x = 25 \). Check: \( \frac{25}{5} - 3 = 5 - 3 = 2 \). Correct. \( x = 25 \); addition property, then multiplication property

8. Solve \( 7 - 2x = 19 \).
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   Subtract 7 from both sides: \( -2x = 12 \). The coefficient is \( -2 \), including the sign. Divide both sides by \( -2 \): \( x = -6 \). Check: \( 7 - 2(-6) = 7 + 12 = 19 \). Correct. \( x = -6 \)

9. A student solves \( 3x - 5 = 16 \) and gets \( x = \frac{11}{3} \). Find their error.
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   They subtracted 5 from the right instead of adding it, getting \( 3x = 11 \). The \( -5 \) is on the left, so to remove it you add 5, and you must add 5 to both sides: \( 3x = 21 \), so \( x = 7 \). Checking the student's answer in the original shows the problem: \( 3(\frac{11}{3}) - 5 = 11 - 5 = 6 \), not 16. \( x = 7 \); they subtracted 5 instead of adding it

10. A taxi charges a $3.50 flag fee plus $2.25 per mile. A ride cost $21.50. Write and solve an equation for the distance, and state the property used to remove the flag fee.
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    Let \( m \) be the distance in miles. The cost is the fixed fee plus the per-mile charge times the distance: \[ 2.25m + 3.50 = 21.50 \] Subtract 3.50 from both sides, which is the subtraction property of equality: \( 2.25m = 18.00 \). Divide both sides by 2.25: \( m = 18 \div 2.25 = 8 \). Check: \( 2.25(8) + 3.50 = 18.00 + 3.50 = 21.50 \). Correct, and 8 miles is a plausible taxi ride. 8 miles; the subtraction property of equality removes the flag fee

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( 5x = 2x + 12 \).
   Show the full solution

   Subtract \( 2x \): \( 3x = 12 \). \( x = 4 \)

2. Solve \( 7x - 4 = 3x + 8 \).
   Show the full solution

   Subtract \( 3x \): \( 4x - 4 = 8 \). Add 4: \( 4x = 12 \). \( x = 3 \)

3. Solve \( 2(x + 5) = 16 \).
   Show the full solution

   Distribute: \( 2x + 10 = 16 \). Subtract 10: \( 2x = 6 \). \( x = 3 \)

4. Solve \( 3(x - 2) = x + 4 \).
   Show the full solution

   Distribute: \( 3x - 6 = x + 4 \). Subtract \( x \): \( 2x - 6 = 4 \). Add 6: \( 2x = 10 \). \( x = 5 \)

5. Solve \( 4x + 3 = 6x - 7 \).
   Show the full solution

   Move the smaller coefficient: subtract \( 4x \) to keep the variable positive on the right. \( 3 = 2x - 7 \). Add 7: \( 10 = 2x \). \( x = 5 \)

6. Solve \( 12 - 3(x - 4) = 2x + 4 \) and check.
   Show the full solution

   The multiplier on the bracket is \( -3 \): \( -3 \times x = -3x \) and \( -3 \times (-4) = +12 \). So the left is \( 12 - 3x + 12 = 24 - 3x \). The equation is \( 24 - 3x = 2x + 4 \). Add \( 3x \) to both sides to keep the variable positive: \( 24 = 5x + 4 \). Subtract 4: \( 20 = 5x \), so \( x = 4 \). Check: left \( = 12 - 3(4 - 4) = 12 - 0 = 12 \); right \( = 2(4) + 4 = 12 \). Correct. \( x = 4 \)

7. Solve \( 2(3x - 1) - 4 = 5(x + 2) \).
   Show the full solution

   Left: \( 6x - 2 - 4 = 6x - 6 \). Right: \( 5x + 10 \). Equation: \( 6x - 6 = 5x + 10 \). Subtract \( 5x \): \( x - 6 = 10 \). Add 6: \( x = 16 \). Check: left \( = 2(47) - 4 = 94 - 4 = 90 \); right \( = 5(18) = 90 \). Correct. \( x = 16 \)

8. Solve \( 8 - (2x + 3) = 5 - x \).
   Show the full solution

   The bare minus in front of the bracket is \( -1 \) times it, so it negates both terms: \( 8 - 2x - 3 = 5 - 2x \). The equation is \( 5 - 2x = 5 - x \). Add \( 2x \) to both sides: \( 5 = 5 + x \). Subtract 5: \( x = 0 \). Check: left \( = 8 - (0 + 3) = 5 \); right \( = 5 - 0 = 5 \). Correct. Zero is a perfectly good solution and is not the same as "no solution". \( x = 0 \)

9. Two gyms compete. Gym A charges $40 to join and $25 per month. Gym B charges $10 to join and $30 per month. After how many months do they cost the same?
   Show the full solution

   Let \( m \) be the number of months. Gym A costs \( 25m + 40 \); Gym B costs \( 30m + 10 \). Setting them equal: \[ 25m + 40 = 30m + 10 \] Subtract \( 25m \) to keep the variable positive: \( 40 = 5m + 10 \). Subtract 10: \( 30 = 5m \), so \( m = 6 \). Check: A costs \( 25(6) + 40 = \$190 \); B costs \( 30(6) + 10 = \$190 \). Equal at 6 months. Before 6 months B is cheaper; after 6 months A is cheaper, because A's monthly rate is lower. 6 months, at $190 each

10. A student solves \( 4(x + 2) = 2(x + 9) \) and gets \( x = 11 \). Find the error and the correct solution.
    Show the full solution

    Correctly: left is \( 4x + 8 \), right is \( 2x + 18 \). Subtract \( 2x \): \( 2x + 8 = 18 \). Subtract 8: \( 2x = 10 \), so \( x = 5 \). Check: left \( = 4(7) = 28 \); right \( = 2(14) = 28 \). Correct. The student most likely distributed only to the first term inside each bracket, getting \( 4x + 2 = 2x + 9 \), which gives \( 2x = 7 \) and not 11, or distributed the 4 but not the 2. Substituting their answer into the original settles it: \( 4(13) = 52 \) but \( 2(20) = 40 \), and \( 52 \neq 40 \). \( x = 5 \); the distribution was incomplete

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \dfrac{x}{3} = 7 \).
   Show the full solution

   Multiply both sides by 3. \( x = 21 \)

2. Solve \( \dfrac{x}{2} + 1 = 6 \).
   Show the full solution

   Multiply every term by 2: \( x + 2 = 12 \). \( x = 10 \)

3. Solve \( 0.5x = 12 \).
   Show the full solution

   Multiply both sides by 2, or divide by 0.5. \( x = 24 \)

4. Solve \( \dfrac{2x}{5} = 4 \).
   Show the full solution

   Multiply by 5: \( 2x = 20 \). \( x = 10 \)

5. Solve \( 0.2x + 1.4 = 3 \).
   Show the full solution

   Multiply every term by 10: \( 2x + 14 = 30 \). Subtract 14: \( 2x = 16 \). \( x = 8 \)

6. Solve \( \dfrac{x}{4} + \dfrac{x}{3} = 14 \).
   Show the full solution

   The least common denominator of 4 and 3 is 12. Multiply every term by 12: \( 3x + 4x = 168 \), so \( 7x = 168 \) and \( x = 24 \). Check: \( \frac{24}{4} + \frac{24}{3} = 6 + 8 = 14 \). Correct. \( x = 24 \)

7. Solve \( \dfrac{3x - 1}{4} = 5 \).
   Show the full solution

   The fraction bar groups the whole numerator, so multiplying both sides by 4 gives \( 3x - 1 = 20 \). Add 1: \( 3x = 21 \), so \( x = 7 \). Check: \( \frac{3(7) - 1}{4} = \frac{20}{4} = 5 \). Correct. \( x = 7 \)

8. Solve \( 0.15x + 0.4 = 0.05x + 1.2 \).
   Show the full solution

   The longest decimal has two places, so multiply every term by 100: \( 15x + 40 = 5x + 120 \). Subtract \( 5x \): \( 10x + 40 = 120 \). Subtract 40: \( 10x = 80 \), so \( x = 8 \). Check: left \( = 0.15(8) + 0.4 = 1.2 + 0.4 = 1.6 \); right \( = 0.05(8) + 1.2 = 0.4 + 1.2 = 1.6 \). Correct. \( x = 8 \)

9. A student solves \( \dfrac{x}{5} + 3 = 8 \) by multiplying by 5 and writes \( x + 3 = 40 \). Find the error.
   Show the full solution

   They multiplied the fraction and the right side by 5 but left the 3 alone. Every term must be multiplied, so it should read \( x + 15 = 40 \), giving \( x = 25 \). Check: \( \frac{25}{5} + 3 = 5 + 3 = 8 \). Correct. Their version gives \( x = 37 \), and substituting that into the original gives \( \frac{37}{5} + 3 = 10.4 \), not 8. \( x = 25 \); the constant term 3 was not multiplied

10. Solve \( \dfrac{x + 2}{3} - \dfrac{x - 1}{4} = 2 \).
    Show the full solution

    The least common denominator of 3 and 4 is 12. Multiply every term by 12, keeping each numerator grouped: \[ 12 \cdot \frac{x + 2}{3} - 12 \cdot \frac{x - 1}{4} = 12 \cdot 2 \] \[ 4(x + 2) - 3(x - 1) = 24 \] Now distribute, taking care with the \( -3 \): \( 4x + 8 - 3x + 3 = 24 \). Combine: \( x + 11 = 24 \), so \( x = 13 \). Check: \( \frac{15}{3} - \frac{12}{4} = 5 - 3 = 2 \). Correct. The brackets around each numerator are essential; dropping them would give \( 4x + 2 - 3x - 1 \) and the wrong answer. \( x = 13 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( x + 4 = x + 9 \).
   Show the full solution

   Subtract \( x \): \( 4 = 9 \), which is false. No solution

2. Solve \( 5x - 2 = 5x - 2 \).
   Show the full solution

   The sides are identical; subtracting gives \( 0 = 0 \), which is true. All real numbers

3. Solve \( 2x + 6 = 2(x + 3) \).
   Show the full solution

   The right expands to \( 2x + 6 \), matching the left. All real numbers

4. Solve \( 4x = 4x + 1 \).
   Show the full solution

   Subtract \( 4x \): \( 0 = 1 \), false. No solution

5. Solve \( 3x + 5 = 3x + 5 \).
   Show the full solution

   All real numbers

6. Solve \( 6(x - 1) = 6x - 6 \) and test two values to confirm.
   Show the full solution

   The left expands to \( 6x - 6 \), identical to the right, so subtracting gives \( 0 = 0 \) and the equation is an identity. Testing: at \( x = 2 \), left \( = 6(1) = 6 \) and right \( = 12 - 6 = 6 \); at \( x = -3 \), left \( = 6(-4) = -24 \) and right \( = -18 - 6 = -24 \). Both work. All real numbers

7. Solve \( 4(x + 2) = 4x + 7 \) and explain what the result means about the two expressions.
   Show the full solution

   The left expands to \( 4x + 8 \), so the equation is \( 4x + 8 = 4x + 7 \). Subtracting \( 4x \) gives \( 8 = 7 \), which is false, so there is no solution. The meaning is that the two expressions differ by exactly 1 no matter what \( x \) is: \( (4x + 8) - (4x + 7) = 1 \) always. Two expressions that always differ by a nonzero amount can never be equal. No solution; the expressions differ by 1 for every \( x \)

8. Solve \( 2(3x - 4) + 5 = 6x - 3 \).
   Show the full solution

   Left: \( 6x - 8 + 5 = 6x - 3 \). The equation is \( 6x - 3 = 6x - 3 \), an identity. Subtracting \( 6x \) and adding 3 gives \( 0 = 0 \). All real numbers

9. A student solves an equation, reaches \( 0 = 0 \), and writes \( x = 0 \). Explain why this is wrong and what the answer should be.
   Show the full solution

   Reaching \( 0 = 0 \) means the variable canceled out and the statement that remains is true regardless of \( x \). That says every value works, not that one particular value does. Writing \( x = 0 \) claims that zero is the only solution, which is the opposite of what was found. The test that settles it is substituting any other value: if \( x = 5 \) also satisfies the original equation, then the answer cannot be a single number. The answer is all real numbers; \( 0 = 0 \) means the equation is always true

10. Find the value of \( k \) that makes \( 5x + k = 5x + 12 \) have infinitely many solutions, and state what happens for every other value of \( k \).
    Show the full solution

    Subtracting \( 5x \) from both sides leaves \( k = 12 \). This is a statement about \( k \), not about \( x \). If \( k \) is 12, the leftover statement is \( 12 = 12 \), which is true, so the equation holds for every \( x \) and there are infinitely many solutions. For any other value of \( k \), the leftover statement is false, so the equation has no solution. Notice there is no value of \( k \) giving exactly one solution: because the coefficients of \( x \) match on both sides, the \( x \) terms always cancel, and the equation can only be always true or never true. \( k = 12 \); every other value gives no solution, and no value gives exactly one

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( d = rt \) for \( t \).
   Show the full solution

   Divide both sides by \( r \), assuming \( r \neq 0 \). \( t = \dfrac{d}{r} \)

2. Solve \( P = 2l + 2w \) for \( w \).
   Show the full solution

   Subtract \( 2l \): \( P - 2l = 2w \). Divide by 2. \( w = \dfrac{P - 2l}{2} \)

3. Solve \( A = lw \) for \( l \).
   Show the full solution

   \( l = \dfrac{A}{w} \)

4. Solve \( y = mx + b \) for \( m \).
   Show the full solution

   Subtract \( b \): \( y - b = mx \). Divide by \( x \), assuming \( x \neq 0 \). \( m = \dfrac{y - b}{x} \)

5. Solve \( C = 2\pi r \) for \( r \).
   Show the full solution

   Divide both sides by \( 2\pi \). \( r = \dfrac{C}{2\pi} \)

6. Solve \( A = \dfrac{1}{2}bh \) for \( h \), and check with \( b = 6 \), \( h = 4 \).
   Show the full solution

   Multiply both sides by 2: \( 2A = bh \). Divide by \( b \): \( h = \frac{2A}{b} \). Check: with \( b = 6 \) and \( h = 4 \), the original gives \( A = \frac{1}{2}(6)(4) = 12 \). The rearranged form gives \( h = \frac{2(12)}{6} = \frac{24}{6} = 4 \). Correct. \( h = \dfrac{2A}{b} \)

7. Solve \( F = \dfrac{9}{5}C + 32 \) for \( C \), and use it to convert 68 degrees Fahrenheit.
   Show the full solution

   Subtract 32: \( F - 32 = \frac{9}{5}C \). Multiply both sides by \( \frac{5}{9} \), which is the reciprocal: \[ C = \frac{5}{9}(F - 32) \] At \( F = 68 \): \( C = \frac{5}{9}(36) = 20 \) degrees Celsius. Check in the original: \( \frac{9}{5}(20) + 32 = 36 + 32 = 68 \). Correct. \( C = \frac{5}{9}(F - 32) \); 68°F is 20°C

8. Solve \( A = \dfrac{1}{2}(b_1 + b_2)h \) for \( b_1 \).
   Show the full solution

   Multiply both sides by 2: \( 2A = (b_1 + b_2)h \). Divide both sides by \( h \): \( \frac{2A}{h} = b_1 + b_2 \). Subtract \( b_2 \): \[ b_1 = \frac{2A}{h} - b_2 \] Check with \( b_1 = 5 \), \( b_2 = 9 \), \( h = 4 \): the original gives \( A = \frac{1}{2}(14)(4) = 28 \). The rearranged form gives \( b_1 = \frac{56}{4} - 9 = 14 - 9 = 5 \). Correct. \( b_1 = \dfrac{2A}{h} - b_2 \)

9. Solve \( S = \dfrac{n}{2}(a + l) \) for \( n \), and state any condition required.
   Show the full solution

   Multiply both sides by 2: \( 2S = n(a + l) \). Divide both sides by the whole quantity \( (a + l) \): \[ n = \frac{2S}{a + l} \] This requires \( a + l \neq 0 \). Check with \( n = 10 \), \( a = 1 \), \( l = 19 \): the original gives \( S = \frac{10}{2}(20) = 100 \), and the rearranged form gives \( n = \frac{200}{20} = 10 \). Correct. \( n = \dfrac{2S}{a + l} \), provided \( a + l \neq 0 \)

10. Solve \( y = \dfrac{x + 3}{x - 1} \) for \( x \), and explain why this one requires the factoring step.
    Show the full solution

    Multiply both sides by \( (x - 1) \), assuming \( x \neq 1 \): \[ y(x - 1) = x + 3 \] Distribute: \( yx - y = x + 3 \). The target now appears in two terms, \( yx \) and \( x \), on opposite sides, so no single division can isolate it. Collect them on one side and the rest on the other: \[ yx - x = 3 + y \] Factor \( x \) out of the left side, which is the step the problem turns on: \[ x(y - 1) = y + 3 \] Divide by \( (y - 1) \), which requires \( y \neq 1 \): \[ x = \frac{y + 3}{y - 1} \] Check with \( x = 5 \): the original gives \( y = \frac{8}{4} = 2 \), and the rearranged form gives \( x = \frac{5}{1} = 5 \). Correct. The factoring was necessary because the target appeared twice; this is the same move as the worked example, in a less obvious setting. \( x = \dfrac{y + 3}{y - 1} \), provided \( y \neq 1 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( x + 5 \lt 12 \).
   Show the full solution

   Subtract 5; direction unchanged. \( x \lt 7 \)

2. Solve \( 3x \geq 21 \).
   Show the full solution

   Divide by 3, a positive number, so no reversal. \( x \geq 7 \)

3. Solve \( -2x \lt 10 \).
   Show the full solution

   Divide by \( -2 \) and reverse the direction. \( x \gt -5 \)

4. Solve \( x - 4 \geq -1 \).
   Show the full solution

   Add 4; direction unchanged. \( x \geq 3 \)

5. Solve \( \dfrac{x}{-4} \gt 2 \).
   Show the full solution

   Multiply by \( -4 \) and reverse. \( x \lt -8 \)

6. Solve \( 5 - 2x \leq 13 \) and check a value.
   Show the full solution

   Subtract 5: \( -2x \leq 8 \). Divide by \( -2 \) and reverse: \( x \geq -4 \). Check inside: at \( x = 0 \), \( 5 - 0 = 5 \leq 13 \), true. Check outside: at \( x = -10 \), \( 5 + 20 = 25 \leq 13 \), false. Correct. \( x \geq -4 \)

7. Solve \( 4(x - 3) \gt 2x + 6 \).
   Show the full solution

   Distribute: \( 4x - 12 \gt 2x + 6 \). Subtract \( 2x \): \( 2x - 12 \gt 6 \). Add 12: \( 2x \gt 18 \). Divide by 2, positive, so no reversal: \( x \gt 9 \). Check at \( x = 10 \): left \( = 4(7) = 28 \), right \( = 26 \), and \( 28 \gt 26 \) is true. \( x \gt 9 \)

8. A student solves \( -x \gt 3 \) and writes \( x \gt -3 \). Find the error and test it.
   Show the full solution

   Dividing both sides by \( -1 \) requires reversing the direction, giving \( x \lt -3 \). Testing the student's answer with \( x = 0 \), which satisfies \( x \gt -3 \): the original becomes \( -0 \gt 3 \), that is \( 0 \gt 3 \), which is false. Testing the correct answer with \( x = -5 \): \( -(-5) = 5 \gt 3 \), true. \( x \lt -3 \); they did not reverse when dividing by a negative

9. A delivery van can carry at most 1,200 kg. It already holds 340 kg of equipment. Each box weighs 26 kg. Write and solve an inequality for the number of boxes it can carry.
   Show the full solution

   Let \( b \) be the number of boxes. The total load must not exceed 1,200 kg: \[ 26b + 340 \leq 1200 \] Subtract 340: \( 26b \leq 860 \). Divide by 26, a positive number, so no reversal: \( b \leq 33.07\ldots \) Boxes come in whole numbers, and the context requires not exceeding the limit, so the answer rounds down: at most 33 boxes. Check: \( 26(33) + 340 = 858 + 340 = 1198 \) kg, within the limit, while 34 boxes would give 1,224 kg, over it. \( 26b + 340 \leq 1200 \), so at most 33 boxes

10. Explain, using the number line, why multiplying both sides of \( 2 \lt 5 \) by \( -1 \) reverses the inequality.
    Show the full solution

    On the number line, \( 2 \lt 5 \) says 2 sits to the left of 5. Multiplying by \( -1 \) sends every number to its mirror image on the opposite side of zero: 2 goes to \( -2 \) and 5 goes to \( -5 \). Reflection reverses left and right, so the number that was further left is now further right. Since 5 was further from zero on the positive side, \( -5 \) is now further from zero on the negative side, which places it to the left of \( -2 \). So \( -2 \gt -5 \). The direction had to reverse because the operation reverses the order of the whole line, not because of any special rule for inequalities. Checking the numbers confirms it: \( -2 \) is indeed greater than \( -5 \). Negating reflects the line across zero, which reverses left and right, so \( 2 \lt 5 \) becomes \( -2 \gt -5 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( |-7| \) and \( |7| \).
   Show the full solution

   Both are 7 units from zero. Both equal 7

2. Solve \( |x| = 6 \).
   Show the full solution

   Two numbers are 6 units from zero. \( x = 6 \) or \( x = -6 \)

3. Solve \( |x| \lt 4 \).
   Show the full solution

   Less than gives and, a single interval. \( -4 \lt x \lt 4 \)

4. Solve \( |x| \geq 3 \).
   Show the full solution

   Greater than gives or, two pieces. \( x \leq -3 \) or \( x \geq 3 \)

5. Solve \( |x| = -5 \).
   Show the full solution

   A distance cannot be negative. No solution

6. Solve \( |x + 3| = 10 \).
   Show the full solution

   Split into two cases: \( x + 3 = 10 \) gives \( x = 7 \), and \( x + 3 = -10 \) gives \( x = -13 \). Check: \( |7 + 3| = 10 \) and \( |-13 + 3| = |-10| = 10 \). Both work. \( x = 7 \) or \( x = -13 \)

7. Solve \( |3x - 1| \leq 8 \).
   Show the full solution

   Less than or equal gives and: \( -8 \leq 3x - 1 \leq 8 \). Add 1 everywhere: \( -7 \leq 3x \leq 9 \). Divide by 3: \( -\frac{7}{3} \leq x \leq 3 \). Check at \( x = 0 \): \( |-1| = 1 \leq 8 \), true. \( -\frac{7}{3} \leq x \leq 3 \)

8. Solve \( 2|x - 4| + 1 = 11 \).
   Show the full solution

   Isolate the absolute value first, before splitting. Subtract 1: \( 2|x - 4| = 10 \). Divide by 2: \( |x - 4| = 5 \). Now split: \( x - 4 = 5 \) gives \( x = 9 \), and \( x - 4 = -5 \) gives \( x = -1 \). Check: \( 2|9 - 4| + 1 = 2(5) + 1 = 11 \), and \( 2|-1 - 4| + 1 = 2(5) + 1 = 11 \). Both work. Splitting before isolating would have produced the wrong two cases. \( x = 9 \) or \( x = -1 \)

9. A machine cuts rods to 250 mm with a tolerance of 3 mm. Write an absolute value inequality for the acceptable lengths and solve it.
   Show the full solution

   A tolerance of 3 mm means the length may differ from 250 by at most 3, in either direction. Distance from the target is exactly what absolute value measures, so with \( L \) the actual length in mm: \[ |L - 250| \leq 3 \] Since this is a "less than or equal", it becomes an and statement: \( -3 \leq L - 250 \leq 3 \). Adding 250 everywhere gives \( 247 \leq L \leq 253 \). A rod of 251 mm is acceptable; one of 254 mm is not. \( |L - 250| \leq 3 \), so \( 247 \leq L \leq 253 \) mm

10. Explain why \( |x| \lt k \) gives one interval while \( |x| \gt k \) gives two, for \( k \gt 0 \).
    Show the full solution

    Read both as statements about distance from zero. "The distance from \( x \) to zero is less than \( k \)" describes every point in the stretch of number line within \( k \) units of the origin. That stretch is connected: it runs from \( -k \) to \( k \) with no gaps, so it is one interval, and both conditions \( x \gt -k \) and \( x \lt k \) must hold at once, which is why it is an and. "The distance from \( x \) to zero is greater than \( k \)" describes every point outside that stretch. But outside a middle section of the line means being off the right-hand end or off the left-hand end, and those are two separate regions with the whole interval between them, so a number satisfies one or the other but never both. That is why it is an or and why the answer has two pieces. The geometry of the number line, not a memorized rule, decides which connector applies. Within \( k \) of zero is a single connected stretch, so and; beyond \( k \) from zero is two separate regions on either side, so or

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Is \( \{(2, 4), (3, 9), (4, 16)\} \) a function?
   Show the full solution

   All inputs differ. Yes

2. Is \( \{(5, 1), (5, 2), (6, 3)\} \) a function?
   Show the full solution

   The input 5 pairs with both 1 and 2. No

3. Is \( \{(1, 7), (2, 7), (3, 7)\} \) a function?
   Show the full solution

   Repeated outputs are allowed; the inputs are all different. Yes

4. State the vertical line test.
   Show the full solution

   If any vertical line crosses the graph more than once, the graph is not a function

5. Is a horizontal line a function?
   Show the full solution

   Every vertical line crosses it exactly once. Yes; it is a constant function

6. Is a vertical line a function? Explain.
   Show the full solution

   A vertical line such as \( x = 4 \) consists of every point with \( x = 4 \) and any \( y \) whatsoever. That single input therefore has infinitely many outputs, which violates the definition immediately. It also fails the vertical line test in the most extreme possible way, since the line \( x = 4 \) coincides with the graph. No; the single input \( x = 4 \) has infinitely many outputs

7. A table shows inputs 1, 2, 3, 2 with outputs 4, 6, 8, 6. Is it a function?
   Show the full solution

   The input 2 appears twice, but both times it is paired with the output 6. The definition forbids one input having two different outputs; repeating the same pair is just redundant information, not a contradiction. So this is a function. Yes; the repeated input has the same output both times

8. Give an example of a relation that is not a function, and change one number to make it one.
   Show the full solution

   \( \{(4, 1), (4, 9), (7, 2)\} \) is not a function, because the input 4 has two different outputs. Changing the first coordinate of either offending pair fixes it: \( \{(4, 1), (5, 9), (7, 2)\} \) is a function, since the inputs are now all different. Changing an output instead, to \( \{(4, 1), (4, 1), (7, 2)\} \), also works, by making the repeated input agree with itself. \( \{(4, 1), (4, 9), (7, 2)\} \) is not; \( \{(4, 1), (5, 9), (7, 2)\} \) is

9. A student says \( y = x^2 \) is not a function because \( x = 2 \) and \( x = -2 \) both give \( y = 4 \). Correct them.
   Show the full solution

   They have applied the test in the wrong direction. The definition restricts outputs per input, not inputs per output. Here \( x = 2 \) gives exactly one output, 4, and \( x = -2 \) gives exactly one output, also 4. Two different inputs sharing an output is entirely permitted, in the same way two people can share a birthday without anyone having two birthdays. The graph passes the vertical line test; what the student noticed is that it fails a horizontal line test, which is a different question with a different meaning. It is a function; repeated outputs are allowed, only repeated inputs with different outputs are not

10. A vending machine is described as a function from buttons to products. Describe a malfunction that would make it stop being a function, and one that would not.
    Show the full solution

    The machine is a function if each button always delivers exactly one determined product. A malfunction that breaks this is a button that sometimes gives a candy bar and sometimes gives a bag of chips: one input, two possible outputs, which is exactly the failure the definition forbids. A malfunction that does not break it is two different buttons both delivering the same product, since sharing an output is allowed and the machine is still perfectly predictable. The distinction is practical: you can rely on a machine of the second kind and not on one of the first, which is why the definition is one-directional. One button giving different products on different presses breaks it; two buttons giving the same product does not

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( f(x) = 3x - 5 \), find \( f(4) \).
   Show the full solution

   \( 3(4) - 5 = 12 - 5 \). 7

2. For \( f(x) = x^2 + 2 \), find \( f(-3) \).
   Show the full solution

   \( (-3)^2 + 2 = 9 + 2 \). 11

3. For \( g(x) = 5 - 2x \), find \( g(0) \).
   Show the full solution

   5

4. For \( f(x) = 4x \), solve \( f(x) = 28 \).
   Show the full solution

   \( 4x = 28 \). \( x = 7 \)

5. If \( f(2) = 9 \), name a point on the graph of \( f \).
   Show the full solution

   \( (2, 9) \)

6. For \( f(x) = x^2 - 4x \), find \( f(5) \) and solve \( f(x) = 0 \).
   Show the full solution

   Evaluating: \( f(5) = 25 - 20 = 5 \). Solving: \( x^2 - 4x = 0 \), so \( x(x - 4) = 0 \), giving \( x = 0 \) or \( x = 4 \). Check: \( f(0) = 0 \) and \( f(4) = 16 - 16 = 0 \). Both work. \( f(5) = 5 \); \( x = 0 \) or \( x = 4 \)

7. For \( f(x) = 2x + 1 \), find \( f(x + 3) \) and simplify.
   Show the full solution

   Substitute the whole of \( x + 3 \) with brackets: \( f(x + 3) = 2(x + 3) + 1 = 2x + 6 + 1 = 2x + 7 \). Check at \( x = 1 \): the result gives \( 2(1) + 7 = 9 \), and \( f(4) = 2(4) + 1 = 9 \). They agree. \( 2x + 7 \)

8. For \( h(x) = x^2 - 1 \), find \( h(a) + h(2) \).
   Show the full solution

   \( h(a) = a^2 - 1 \) and \( h(2) = 4 - 1 = 3 \), so the sum is \( a^2 - 1 + 3 = a^2 + 2 \). Note this is not the same as \( h(a + 2) \), which would be \( (a + 2)^2 - 1 = a^2 + 4a + 3 \). Adding outputs and adding inputs are different operations, and checking at \( a = 1 \) confirms it: the first gives 3, the second gives 8. \( a^2 + 2 \)

9. A student writes that \( f(3) \) means \( f \) times 3. Explain the error and give a case where the distinction changes the answer.
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   \( f \) is the name of a rule, not a number, so there is nothing to multiply by. The brackets indicate which input the rule is being applied to. Take \( f(x) = x + 10 \). Then \( f(3) = 13 \), the output at input 3. Reading it as multiplication suggests some quantity times 3, which is not 13 for any sensible interpretation of \( f \) as a number. The confusion is real because the notation genuinely is ambiguous in appearance, which is why function names are single letters and why context has to settle it. \( f \) names a rule, not a factor; \( f(3) \) is the output when the input is 3

10. A taxi fare is \( C(m) = 2.25m + 3.50 \) dollars for \( m \) miles. Find \( C(8) \), solve \( C(m) = 30.50 \), and state what each answer means in context.
    Show the full solution

    Evaluating: \( C(8) = 2.25(8) + 3.50 = 18.00 + 3.50 = \$21.50 \). This is the cost of an 8 mile ride. Solving: \( 2.25m + 3.50 = 30.50 \), so \( 2.25m = 27.00 \) and \( m = 27 \div 2.25 = 12 \) miles. This is the distance you could travel for $30.50. Check: \( 2.25(12) + 3.50 = 27.00 + 3.50 = 30.50 \). Correct. The two questions face opposite directions: one supplies a distance and asks for a cost, the other supplies a cost and asks for a distance, and confusing them is the commonest setup error on modeling problems. \( C(8) = \$21.50 \); \( m = 12 \) miles

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the domain of \( f(x) = 4x + 1 \).
   Show the full solution

   A polynomial has no restrictions. All real numbers

2. State the domain of \( f(x) = \dfrac{1}{x} \).
   Show the full solution

   The denominator is zero at \( x = 0 \). All real numbers except 0

3. State the domain of \( f(x) = \sqrt{x} \).
   Show the full solution

   The radicand must be nonnegative. \( x \geq 0 \)

4. State the range of \( f(x) = x^2 \).
   Show the full solution

   A square is never negative and every nonnegative value is achieved. \( y \geq 0 \)

5. State the domain of \( f(x) = \dfrac{3}{x + 2} \).
   Show the full solution

   \( x + 2 = 0 \) at \( x = -2 \). All real numbers except \( -2 \)

6. State the domain of \( f(x) = \sqrt{10 - 2x} \).
   Show the full solution

   Require \( 10 - 2x \geq 0 \). Subtract 10: \( -2x \geq -10 \). Divide by \( -2 \) and reverse the inequality: \( x \leq 5 \). Check at \( x = 5 \): \( \sqrt{0} = 0 \), defined. At \( x = 6 \): \( \sqrt{-2} \), undefined. Correct. \( x \leq 5 \)

7. State the range of \( f(x) = x^2 + 3 \) and justify it.
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   The term \( x^2 \) is never negative and takes every value from 0 upward. Adding 3 shifts all of those up by 3, so the outputs run from 3 upward. The smallest output is 3, reached at \( x = 0 \), and there is no largest. \( y \geq 3 \)

8. A rectangle has perimeter 40 cm and width \( w \). Write its area as a function of \( w \) and state the domain in context.
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   Perimeter gives \( 2w + 2l = 40 \), so \( l = 20 - w \). The area is \( A(w) = w(20 - w) = 20w - w^2 \). Algebraically this polynomial accepts every real number, but in context both dimensions must be positive: \( w \gt 0 \) and \( 20 - w \gt 0 \), the second giving \( w \lt 20 \). So the domain is \( 0 \lt w \lt 20 \) cm. At the excluded endpoints the rectangle would have zero width or zero length and would not be a rectangle. \( A(w) = 20w - w^2 \), with \( 0 \lt w \lt 20 \)

9. A student gives the domain of \( f(x) = \dfrac{x - 3}{x - 3} \) as all real numbers, because it simplifies to 1. Explain the error.
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   The simplification is valid only where the cancellation is legitimate, and dividing \( (x - 3) \) by itself requires that it not be zero. At \( x = 3 \) the original expression is \( \frac{0}{0} \), which is undefined, so 3 is not in the domain. The simplified expression \( f(x) = 1 \) agrees with the original everywhere except at that one point, where the original has a hole. The domain is determined by the function as given, before simplification, which is why excluded values must be recorded at the start. All real numbers except 3; the domain is set by the original form, not the simplified one

10. A ball is thrown and its height in meters after \( t \) seconds is \( h(t) = -5t^2 + 20t \). State the domain and range in context.
    Show the full solution

    Algebraically the polynomial accepts all real \( t \), but the context restricts it. Time starts at the throw, so \( t \geq 0 \), and the model stops being meaningful once the ball lands. It lands when \( h = 0 \): \( -5t^2 + 20t = 0 \), so \( -5t(t - 4) = 0 \), giving \( t = 0 \) or \( t = 4 \). The domain in context is therefore \( 0 \leq t \leq 4 \) seconds. For the range, the graph is a downward parabola with zeros at 0 and 4, so its maximum is halfway between them at \( t = 2 \): \( h(2) = -5(4) + 40 = -20 + 40 = 20 \) meters. The height runs from 0 up to 20 and back to 0, so the range is \( 0 \leq h \leq 20 \) meters. Domain \( 0 \leq t \leq 4 \) s; range \( 0 \leq h \leq 20 \) m

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. How do you find the \( y \) intercept of a function from its equation?
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   Set \( x = 0 \) and evaluate

2. How do you find the \( x \) intercepts?
   Show the full solution

   Set \( y = 0 \) and solve for \( x \)

3. Find the \( y \) intercept of \( f(x) = 2x - 9 \).
   Show the full solution

   \( f(0) = -9 \). \( (0, -9) \)

4. Find the \( x \) intercept of \( f(x) = 3x + 12 \).
   Show the full solution

   \( 3x + 12 = 0 \), so \( x = -4 \). \( (-4, 0) \)

5. Another name for an \( x \) intercept is what?
   Show the full solution

   A zero, or a root

6. For \( f(x) = x^2 - 6x + 5 \), find both intercepts.
   Show the full solution

   \( y \) intercept: \( f(0) = 5 \), so \( (0, 5) \). \( x \) intercepts: \( x^2 - 6x + 5 = 0 \) factors as \( (x - 1)(x - 5) = 0 \), since \( -1 \) and \( -5 \) multiply to 5 and add to \( -6 \). So \( x = 1 \) and \( x = 5 \). \( (0, 5) \); \( (1, 0) \) and \( (5, 0) \)

7. A parabola has zeros at \( x = -2 \) and \( x = 6 \) and opens upward. State the axis of symmetry and whether the vertex is a maximum or minimum.
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   The axis of symmetry is halfway between the zeros: \( x = \frac{-2 + 6}{2} = 2 \). An upward parabola has its lowest point at the vertex, so the vertex is a minimum, located at \( x = 2 \). Axis of symmetry \( x = 2 \); the vertex is a minimum

8. A graph rises from \( (-4, 1) \) to \( (0, 7) \), then falls to \( (5, -3) \). State the intervals of increase and decrease and the maximum.
   Show the full solution

   Intervals are stated in \( x \) values. The graph rises as \( x \) goes from \( -4 \) to 0, so it is increasing on \( -4 \lt x \lt 0 \). It falls as \( x \) goes from 0 to 5, so it is decreasing on \( 0 \lt x \lt 5 \). The turning point is at \( x = 0 \), where the height is 7, so the maximum value is 7 and it occurs at \( (0, 7) \). Increasing on \( -4 \lt x \lt 0 \), decreasing on \( 0 \lt x \lt 5 \); maximum value 7 at \( (0, 7) \)

9. A student says a function is "increasing from 2 to 9" when the graph rises from \( (2, 5) \) to \( (9, 11) \). Correct the statement precisely.
   Show the full solution

   The statement happens to use the right numbers but for an accidental reason, and it is ambiguous as written: a reader cannot tell whether 2 and 9 are inputs or outputs. The interval of increase is a set of inputs, so it must be stated as \( 2 \lt x \lt 9 \). The corresponding outputs run from 5 to 11 and are not what the question asks for. Naming the variable removes all ambiguity, which is why the convention exists. Increasing on the interval \( 2 \lt x \lt 9 \); intervals are stated in \( x \) values

10. The height of a drone is \( h(t) = -2t^2 + 12t \) meters after \( t \) minutes. Find when it is at its highest, that height, and when it returns to the ground, interpreting each.
    Show the full solution

    Zeros first: \( -2t^2 + 12t = 0 \) factors as \( -2t(t - 6) = 0 \), giving \( t = 0 \) and \( t = 6 \). The drone leaves the ground at \( t = 0 \) and returns at \( t = 6 \) minutes. The parabola opens downward, so the vertex is a maximum, halfway between the zeros at \( t = \frac{0 + 6}{2} = 3 \) minutes. The height there is \( h(3) = -2(9) + 36 = -18 + 36 = 18 \) meters. Interpreting: the drone climbs for the first 3 minutes, reaching a greatest height of 18 meters, then descends for 3 minutes and lands 6 minutes after takeoff. It is increasing on \( 0 \lt t \lt 3 \) and decreasing on \( 3 \lt t \lt 6 \), and the sensible domain is \( 0 \leq t \leq 6 \). Highest at \( t = 3 \) min, 18 m; back on the ground at \( t = 6 \) min

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write the formula for average rate of change from \( a \) to \( b \).
   Show the full solution

   \( \dfrac{f(b) - f(a)}{b - a} \)

2. For \( f(x) = 3x + 1 \), find the average rate of change from 0 to 4.
   Show the full solution

   \( f(0) = 1 \), \( f(4) = 13 \). \( \frac{13 - 1}{4} = 3 \). 3

3. For \( f(x) = x^2 \), find the average rate of change from 0 to 3.
   Show the full solution

   \( \frac{9 - 0}{3} \). 3

4. For \( f(x) = 10 - 2x \), find the average rate of change from 1 to 6.
   Show the full solution

   \( f(1) = 8 \), \( f(6) = -2 \). \( \frac{-2 - 8}{5} = \frac{-10}{5} \). \( -2 \)

5. What does a negative average rate of change mean?
   Show the full solution

   The output decreased, on average, across the interval

6. For \( f(x) = x^2 \), find the average rate of change from 3 to 5, and explain why it differs from the answer over 0 to 3.
   Show the full solution

   \( f(3) = 9 \), \( f(5) = 25 \), so \( \frac{25 - 9}{5 - 3} = \frac{16}{2} = 8 \). Over 0 to 3 the answer was 3. They differ because \( x^2 \) is not linear: the curve is much steeper further from the origin, so the same width of interval produces a much larger rise. A parabola has no single rate of change, which is exactly why the word "average" and the interval both have to be specified. 8; the function is not linear, so its steepness varies

7. A town's population was 12,400 in 2010 and 16,900 in 2020. Find the average rate of change and state it as a sentence.
   Show the full solution

   \( \frac{16900 - 12400}{2020 - 2010} = \frac{4500}{10} = 450 \). The units are people per year. 450 people per year: the population grew by an average of 450 people each year over that decade

8. A student computes the average rate of change of \( f \) from 2 to 7 as \( \frac{f(7) - f(2)}{2 - 7} \) and gets \( -3 \). What is the correct answer and what went wrong?
   Show the full solution

   They subtracted the inputs in the opposite order from the outputs. The denominator should be \( 7 - 2 = 5 \), not \( 2 - 7 = -5 \). Dividing by \( -5 \) instead of 5 flips the sign, so the correct answer is \( +3 \). The magnitude is right and the meaning is exactly reversed: the function was increasing, and the student's answer says it was decreasing. \( +3 \); the input difference was taken in the reverse order

9. A function has \( f(2) = 10 \) and \( f(8) = 10 \). Explain why the average rate of change is 0 and why this does not mean the function is constant.
   Show the full solution

   The average rate of change is \( \frac{10 - 10}{8 - 2} = \frac{0}{6} = 0 \). This says only that the function finished the interval at exactly the height it started. It says nothing about the journey. The function could have risen to 50 and come back down, or dropped to \( -20 \) and returned, or genuinely stayed at 10 throughout, and all three give the same average rate of change. The measure compares two endpoints and is blind to everything between them, which is both its convenience and its limitation. 0, because the endpoints have the same output; the function could have varied greatly in between

10. A ball's height is \( h(t) = -5t^2 + 30t \) meters. Find the average rate of change on \( 0 \leq t \leq 2 \) and on \( 4 \leq t \leq 6 \), and explain the signs in context.
    Show the full solution

    First interval: \( h(0) = 0 \) and \( h(2) = -5(4) + 60 = -20 + 60 = 40 \). \( \frac{40 - 0}{2 - 0} = 20 \) meters per second. Second interval: \( h(4) = -5(16) + 120 = -80 + 120 = 40 \) and \( h(6) = -5(36) + 180 = -180 + 180 = 0 \). \( \frac{0 - 40}{6 - 4} = \frac{-40}{2} = -20 \) meters per second. In context, the ball rose at an average of 20 m/s during the first two seconds and fell at an average of 20 m/s during the last two. The signs report direction: positive means rising, negative means falling. The equal magnitudes reflect the symmetry of the parabola about its vertex at \( t = 3 \), where the ball is momentarily at its highest and is neither rising nor falling. 20 m/s rising, then \( -20 \) m/s falling; the sign gives the direction

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning. Use \( f(x) = \begin{cases} x + 4 & \text{if } x \leq 0 \\ 3x & \text{if } x \gt 0 \end{cases} \) where a problem refers to \( f \).

1. Find \( f(-2) \).
   Show the full solution

   \( -2 \leq 0 \), so use the first rule: \( -2 + 4 \). 2

2. Find \( f(5) \).
   Show the full solution

   \( 5 \gt 0 \), so use the second rule: \( 3(5) \). 15

3. Find \( f(0) \).
   Show the full solution

   The first condition is \( x \leq 0 \), which includes 0. 4

4. Which rule applies when \( x = 0.5 \)?
   Show the full solution

   The second, \( 3x \), since \( 0.5 \gt 0 \)

5. Why must the two conditions not overlap?
   Show the full solution

   An overlapping input would have two outputs, so it would not be a function

6. Does the graph of \( f \) have a jump at \( x = 0 \)? Justify.
   Show the full solution

   At \( x = 0 \) the value is \( f(0) = 4 \), from the first piece. Approaching 0 from the right, the second piece \( 3x \) approaches \( 3(0) = 0 \). Since 4 and 0 differ, there is a jump of 4 units. The point \( (0, 4) \) is closed because the first condition includes 0; the point \( (0, 0) \) is open because the second condition excludes it. Yes, a jump of 4 units, with a closed circle at \( (0, 4) \) and an open circle at \( (0, 0) \)

7. Solve \( f(x) = 12 \) for the function above.
   Show the full solution

   Both pieces must be checked, and each answer must satisfy its own condition. First piece: \( x + 4 = 12 \) gives \( x = 8 \). But the first piece requires \( x \leq 0 \), and 8 does not satisfy that, so this solution is rejected. Second piece: \( 3x = 12 \) gives \( x = 4 \), and \( 4 \gt 0 \) is satisfied, so this one is valid. Checking every candidate against its own condition is essential; skipping it is how students report solutions the function never actually produces. \( x = 4 \) only

8. A parking garage charges $5 for up to 2 hours and $3 for each additional hour or part of one. Write a piecewise rule for the cost of \( t \) hours, for \( 0 \lt t \leq 5 \).
   Show the full solution

   For any stay up to and including 2 hours the charge is a flat $5. Beyond 2 hours the $5 still applies and each extra hour, or part of one, adds $3. Treating the extra time as whole hours for a simple model: \[ C(t) = \begin{cases} 5 & \text{if } 0 \lt t \leq 2 \\ 5 + 3(t - 2) & \text{if } 2 \lt t \leq 5 \end{cases} \] Check at \( t = 2 \): the first rule gives $5. At \( t = 3 \): the second gives \( 5 + 3(1) = \$8 \). At \( t = 5 \): \( 5 + 3(3) = \$14 \). The boundary at 2 belongs to the first piece only, which matches "up to 2 hours". \( C(t) = 5 \) for \( 0 \lt t \leq 2 \), and \( 5 + 3(t - 2) \) for \( 2 \lt t \leq 5 \)

9. A student evaluates \( f(-3) \) using the rule \( 3x \) and gets \( -9 \). Find the error.
   Show the full solution

   They substituted before checking the condition. The rule \( 3x \) applies only when \( x \gt 0 \), and \( -3 \) is not greater than 0, so that rule does not apply at all. The condition \( x \leq 0 \) is the one \( -3 \) satisfies, so the correct rule is \( x + 4 \), giving \( f(-3) = -3 + 4 = 1 \). The habit that prevents this is reading the conditions first and the rules second. \( f(-3) = 1 \); they used the rule for positive inputs on a negative input

10. Write a piecewise function whose graph is continuous at \( x = 3 \), using a linear rule below 3 and a linear rule above it, and show that it is continuous.
    Show the full solution

    Continuity at the boundary requires the two pieces to meet, that is, to give the same value at \( x = 3 \). Pick the first piece as \( 2x + 1 \), which at \( x = 3 \) gives 7. The second piece must also give 7 at \( x = 3 \), so choose a different slope and adjust the constant: with slope \( -1 \), require \( -1(3) + c = 7 \), giving \( c = 10 \). So \[ f(x) = \begin{cases} 2x + 1 & \text{if } x \lt 3 \\ -x + 10 & \text{if } x \geq 3 \end{cases} \] Checking: approaching from the left, \( 2x + 1 \) approaches \( 7 \); at \( x = 3 \) the value is \( -3 + 10 = 7 \). The two agree, so there is no jump and the graph is continuous, with a corner rather than a break, since the slopes differ. Many other answers work; the requirement is only that the two rules agree at the boundary. Any pair agreeing at \( x = 3 \), such as \( 2x + 1 \) for \( x \lt 3 \) and \( -x + 10 \) for \( x \geq 3 \), both giving 7 there

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the vertex of \( y = |x| \).
   Show the full solution

   \( (0, 0) \)

2. State the vertex of \( y = |x - 4| \).
   Show the full solution

   The inside is zero at \( x = 4 \). \( (4, 0) \)

3. State the vertex of \( y = |x| + 6 \).
   Show the full solution

   \( (0, 6) \)

4. Which way does \( y = -|x| \) open?
   Show the full solution

   The coefficient is negative. Downward

5. State the vertex of \( y = |x + 2| - 7 \).
   Show the full solution

   The inside is zero at \( x = -2 \). \( (-2, -7) \)

6. Describe how \( y = 3|x - 1| + 2 \) differs from the parent graph.
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   The vertex moves from the origin to \( (1, 2) \), that is, 1 unit right and 2 units up. Since \( a = 3 \) is positive, it still opens upward, and since \( 3 \gt 1 \) it is narrower than the parent, with arms of slope \( \pm 3 \). The minimum value is 2, so the range is \( y \geq 2 \). Shifted right 1 and up 2, opening upward, three times narrower

7. Find the \( x \) intercepts of \( y = |x - 2| - 5 \).
   Show the full solution

   Set \( y = 0 \) and isolate the absolute value: \( |x - 2| = 5 \). Split: \( x - 2 = 5 \) gives \( x = 7 \), and \( x - 2 = -5 \) gives \( x = -3 \). Check: \( |7 - 2| - 5 = 0 \) and \( |-3 - 2| - 5 = 5 - 5 = 0 \). Both correct, and they are symmetric about the vertex at \( x = 2 \). \( x = 7 \) and \( x = -3 \)

8. Write the equation of an absolute value function with vertex \( (-4, 3) \) opening downward with arms of slope 2 in magnitude.
   Show the full solution

   Vertex \( (h, k) = (-4, 3) \), so \( h = -4 \) and \( k = 3 \). Since \( x - h = x - (-4) = x + 4 \), the inside is \( x + 4 \). Opening downward with arm slope magnitude 2 means \( a = -2 \). \[ y = -2|x + 4| + 3 \] Check the vertex: at \( x = -4 \), \( y = -2(0) + 3 = 3 \). Correct. \( y = -2|x + 4| + 3 \)

9. Explain why \( y = |x - 5| \) shifts right rather than left, despite the minus sign.
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   The vertex of an absolute value graph is wherever the expression inside the bars equals zero, because that is where the output is smallest. Setting \( x - 5 = 0 \) gives \( x = 5 \), so the vertex sits at \( x = 5 \), which is 5 units to the right of the origin. Another way to see it: to get the same output the parent function had at input 0, this function needs an input of 5, so every feature occurs 5 units later along the axis, which is a shift right. The sign in the equation is not the direction of the shift; it is part of the expression whose zero locates the vertex. The vertex is where the inside is zero, which is \( x = 5 \), so the graph shifts 5 right

10. A shipping company's charge is $4 plus $1.50 for every kilogram a package differs from 10 kg, in either direction. Write this as an absolute value function and find the cost of a 6 kg and a 13 kg package.
    Show the full solution

    "Differs from 10 in either direction" is exactly a distance from 10, which is \( |w - 10| \) for a weight \( w \) in kilograms. Each kilogram of difference costs $1.50, on top of the $4 base: \[ C(w) = 1.5|w - 10| + 4 \] At \( w = 6 \): \( 1.5|6 - 10| + 4 = 1.5(4) + 4 = 6 + 4 = \$10 \). At \( w = 13 \): \( 1.5|13 - 10| + 4 = 1.5(3) + 4 = 4.5 + 4 = \$8.50 \). The graph is a V with vertex at \( (10, 4) \), so a 10 kg package is cheapest at $4, and the cost rises symmetrically for packages lighter or heavier than that. Absolute value is the natural model whenever a cost or penalty depends on how far a quantity is from a target, regardless of direction. \( C(w) = 1.5|w - 10| + 4 \); $10 and $8.50

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the slope through \( (1, 2) \) and \( (3, 8) \).
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   \( \frac{8 - 2}{3 - 1} = \frac{6}{2} \). 3

2. Find the slope through \( (0, 5) \) and \( (4, 5) \).
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   \( \frac{5 - 5}{4 - 0} = \frac{0}{4} \). The line is horizontal. 0

3. Find the slope through \( (2, 3) \) and \( (2, 9) \).
   Show the full solution

   \( \frac{9 - 3}{2 - 2} = \frac{6}{0} \), which is undefined. The line is vertical. Undefined

4. Find the slope through \( (-1, 4) \) and \( (3, -4) \).
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   \( \frac{-4 - 4}{3 - (-1)} = \frac{-8}{4} \). \( -2 \)

5. What does a slope of 5 mean about the output?
   Show the full solution

   The output increases by 5 for every 1 unit increase in the input

6. A table gives \( (0, 3), (2, 9), (5, 18) \). Is the relationship linear? Justify.
   Show the full solution

   Test the slope between successive pairs. From the first to the second: \( \frac{9 - 3}{2 - 0} = \frac{6}{2} = 3 \). From the second to the third: \( \frac{18 - 9}{5 - 2} = \frac{9}{3} = 3 \). The slopes agree, so the data lies on a single line and the relationship is linear with slope 3. Yes; both pairs give a slope of 3

7. A table gives \( (1, 4), (2, 8), (3, 16) \). Is the relationship linear? Justify.
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   From the first pair: \( \frac{8 - 4}{2 - 1} = 4 \). From the second: \( \frac{16 - 8}{3 - 2} = 8 \). The slopes disagree, so no single line passes through all three points and the relationship is not linear. Notice the outputs are doubling each time, which is the signature of an exponential pattern rather than a linear one. No; the slopes are 4 and 8, so it is not linear

8. A candle burns from 30 cm to 18 cm in 4 hours. Find the rate of change and state it in context.
   Show the full solution

   Treat time as the input and height as the output: the points are \( (0, 30) \) and \( (4, 18) \). \( m = \frac{18 - 30}{4 - 0} = \frac{-12}{4} = -3 \). The units are centimeters per hour. \( -3 \) cm per hour: the candle shortens by 3 cm each hour

9. A student finds the slope through \( (5, 2) \) and \( (1, 10) \) as \( +2 \). Find the error.
   Show the full solution

   Correctly: \( \frac{10 - 2}{1 - 5} = \frac{8}{-4} = -2 \). The student most likely computed \( \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2 \), subtracting the \( y \) values in one order and the \( x \) values in the other. A quick sanity check settles it: as \( x \) goes from 1 to 5 the \( y \) value goes from 10 down to 2, so the line falls and the slope must be negative. \( -2 \); the coordinates were subtracted in inconsistent orders

10. Explain why a vertical line has undefined slope rather than a slope of zero, and what each of the two cases looks like.
    Show the full solution

    Slope is a change in \( y \) divided by a change in \( x \). On a horizontal line the \( y \) values never change, so the numerator is zero while the denominator is some nonzero number, and zero divided by a nonzero number is zero. That is a genuine slope of 0, and it correctly says the output does not change as the input increases. On a vertical line the \( x \) values never change, so the denominator is zero. No number multiplied by zero gives a nonzero numerator, so the quotient does not exist at all. It is not that the slope is zero; it is that the question "how much does \( y \) change per unit increase in \( x \)" has no answer, because \( x \) never increases. A vertical line is also not a function, for the reason given in lesson 3.1. Horizontal lines have a zero numerator, giving slope 0; vertical lines have a zero denominator, so the quotient is undefined

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the slope and \( y \) intercept of \( y = 4x - 3 \).
   Show the full solution

   Slope 4, \( y \) intercept \( (0, -3) \)

2. State the slope and \( y \) intercept of \( y = -x + 7 \).
   Show the full solution

   The coefficient of \( x \) is \( -1 \). Slope \( -1 \), \( y \) intercept \( (0, 7) \)

3. Write the equation of the line with slope 2 and \( y \) intercept \( (0, -6) \).
   Show the full solution

   \( y = 2x - 6 \)

4. State the slope of \( y = 9 \).
   Show the full solution

   There is no \( x \) term, so the coefficient of \( x \) is 0 and the line is horizontal. 0

5. Write \( 2y = 6x + 8 \) in slope-intercept form.
   Show the full solution

   Divide every term by 2. \( y = 3x + 4 \)

6. Write \( 4x + 2y = 10 \) in slope-intercept form and state the slope.
   Show the full solution

   Subtract \( 4x \) from both sides: \( 2y = -4x + 10 \). Divide every term by 2: \( y = -2x + 5 \). The slope is \( -2 \), not 4; reading it off before rearranging would have given the wrong sign and the wrong magnitude. \( y = -2x + 5 \), slope \( -2 \)

7. A line passes through \( (0, -4) \) and has slope \( \frac{3}{5} \). Write its equation and find a second point on it.
   Show the full solution

   The \( y \) intercept is given directly, so \( y = \frac{3}{5}x - 4 \). For a second point, step from \( (0, -4) \) using the slope as rise over run: right 5, up 3, reaching \( (5, -1) \). Check: \( \frac{3}{5}(5) - 4 = 3 - 4 = -1 \). Correct. \( y = \frac{3}{5}x - 4 \); \( (5, -1) \) is on it

8. A phone plan costs $20 per month plus $0.05 per minute. Write the cost function and interpret both parameters.
   Show the full solution

   Let \( t \) be the number of minutes used in a month and \( C \) the cost in dollars. \[ C(t) = 0.05t + 20 \] The intercept 20 means the cost is $20 when no minutes are used, which is the fixed monthly fee. The slope 0.05 means each additional minute adds 5 cents to the bill. Checking at \( t = 200 \): \( C = 10 + 20 = \$30 \), which is sensible. \( C(t) = 0.05t + 20 \); $20 fixed fee, 5 cents per minute

9. A student says the line \( y = 5 \) has undefined slope because there is no \( x \). Correct them.
   Show the full solution

   They have confused horizontal with vertical. Writing \( y = 5 \) as \( y = 0x + 5 \) makes it clear: the coefficient of \( x \) is 0, so the slope is 0 and the line is horizontal at height 5. Whatever \( x \) is, \( y \) stays at 5, which is exactly what a slope of zero means. Undefined slope belongs to vertical lines such as \( x = 5 \), where the denominator of the slope formula is zero. The slope is 0; \( y = 5 \) is horizontal, while \( x = 5 \) would be vertical with undefined slope

10. Two printing companies quote jobs. Company A charges \( y = 0.25x + 40 \) and Company B charges \( y = 0.40x + 10 \), where \( x \) is the number of pages. Find where they cost the same and advise which to choose.
    Show the full solution

    Set the two costs equal: \[ 0.25x + 40 = 0.40x + 10 \] Subtract \( 0.25x \): \( 40 = 0.15x + 10 \). Subtract 10: \( 30 = 0.15x \). Divide: \( x = 30 \div 0.15 = 200 \) pages. Check: A costs \( 0.25(200) + 40 = 50 + 40 = \$90 \); B costs \( 0.40(200) + 10 = 80 + 10 = \$90 \). Equal. To advise, compare on either side. At 100 pages: A is \( \$65 \), B is \( \$50 \), so B is cheaper. At 400 pages: A is \( \$140 \), B is \( \$170 \), so A is cheaper. The reason is in the parameters: B has the lower fixed cost but the higher per-page rate, so it wins on small jobs and loses on large ones. Equal at 200 pages, $90 each; choose B below 200 pages and A above 200

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Write the point-slope equation through \( (3, 5) \) with slope 4.
   Show the full solution

   \( y - 5 = 4(x - 3) \)

2. Write the point-slope equation through \( (0, 2) \) with slope \( -3 \).
   Show the full solution

   \( y - 2 = -3(x - 0) \), or \( y - 2 = -3x \)

3. Write the point-slope equation through \( (-4, 1) \) with slope 2.
   Show the full solution

   \( x - (-4) = x + 4 \). \( y - 1 = 2(x + 4) \)

4. Convert \( y - 3 = 5(x - 1) \) to slope-intercept form.
   Show the full solution

   Distribute: \( y - 3 = 5x - 5 \). Add 3. \( y = 5x - 2 \)

5. What two pieces of information does point-slope form require?
   Show the full solution

   A slope and any one point on the line

6. Find the equation of the line through \( (1, -2) \) and \( (5, 6) \) in slope-intercept form.
   Show the full solution

   Slope: \( \frac{6 - (-2)}{5 - 1} = \frac{8}{4} = 2 \). Using \( (1, -2) \): \( y - (-2) = 2(x - 1) \), that is \( y + 2 = 2x - 2 \), so \( y = 2x - 4 \). Check with the unused point: \( 2(5) - 4 = 6 \). Correct. \( y = 2x - 4 \)

7. Find the equation of the line through \( (-2, 5) \) and \( (4, -7) \).
   Show the full solution

   Slope: \( \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2 \). Using \( (-2, 5) \): \( y - 5 = -2(x + 2) \), so \( y - 5 = -2x - 4 \) and \( y = -2x + 1 \). Check with \( (4, -7) \): \( -2(4) + 1 = -8 + 1 = -7 \). Correct. \( y = -2x + 1 \)

8. A student writes the line through \( (3, -4) \) with slope 5 as \( y - 4 = 5(x - 3) \). Find the error.
   Show the full solution

   The \( y \) coordinate is \( -4 \), so the left side is \( y - (-4) = y + 4 \), not \( y - 4 \). The correct equation is \( y + 4 = 5(x - 3) \), which in slope-intercept form is \( y = 5x - 19 \). Checking the student's version at \( x = 3 \) gives \( y = 4 \), not \( -4 \), so it passes through the wrong point. Substituting the given point into your own answer is the check that catches this instantly. \( y + 4 = 5(x - 3) \); the negative \( y \) coordinate was not subtracted

9. A plumber's fee is linear in hours. A 3 hour job costs $209 and a 7 hour job costs $401. Find the equation and the fixed callout fee.
   Show the full solution

   Treat hours as input and cost as output, giving the points \( (3, 209) \) and \( (7, 401) \). Slope: \( \frac{401 - 209}{7 - 3} = \frac{192}{4} = 48 \) dollars per hour. Point-slope with \( (3, 209) \): \( C - 209 = 48(h - 3) \), so \( C - 209 = 48h - 144 \) and \( C = 48h + 65 \). The fixed callout fee is the cost at zero hours, the intercept: $65. Check with the other point: \( 48(7) + 65 = 336 + 65 = 401 \). Correct. \( C = 48h + 65 \); the callout fee is $65

10. Derive point-slope form from the slope formula, and explain why the result is a complete equation of the line even before rearranging.
    Show the full solution

    Let \( (x_1, y_1) \) be a known point on the line and \( (x, y) \) be any other point on it. By the definition of slope, the slope between them is \[ m = \frac{y - y_1}{x - x_1} \] Multiplying both sides by \( x - x_1 \), which is legitimate for any point other than the known one, gives \[ m(x - x_1) = y - y_1 \] which rearranged is point-slope form. It is a complete equation of the line because it states exactly the condition for a point to be on the line: the slope from the known point to it equals \( m \). Any \( (x, y) \) satisfying it lies on the line, and any point on the line satisfies it. Rearranging into \( y = mx + b \) changes the appearance but not the set of points described, which is why no further work is required unless a question asks for a particular form. It follows from multiplying the slope formula through by \( x - x_1 \), and it states the exact condition for a point to lie on the line

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the \( x \) intercept of \( 2x + 5y = 10 \).
   Show the full solution

   Set \( y = 0 \): \( 2x = 10 \). \( (5, 0) \)

2. Find the \( y \) intercept of \( 2x + 5y = 10 \).
   Show the full solution

   Set \( x = 0 \): \( 5y = 10 \). \( (0, 2) \)

3. Find both intercepts of \( x - 3y = 6 \).
   Show the full solution

   \( y = 0 \) gives \( x = 6 \). \( x = 0 \) gives \( -3y = 6 \), so \( y = -2 \). \( (6, 0) \) and \( (0, -2) \)

4. Write \( y = 2x + 5 \) in standard form.
   Show the full solution

   Subtract \( 2x \): \( -2x + y = 5 \). Multiply by \( -1 \) to make \( A \) positive. \( 2x - y = -5 \)

5. What is the slope of \( 4x + 2y = 9 \)?
   Show the full solution

   Rearranging: \( 2y = -4x + 9 \), so \( y = -2x + 4.5 \). \( -2 \)

6. Find both intercepts and the slope of \( 5x - 2y = 20 \).
   Show the full solution

   \( x \) intercept: set \( y = 0 \), giving \( 5x = 20 \) and \( x = 4 \), so \( (4, 0) \). \( y \) intercept: set \( x = 0 \), giving \( -2y = 20 \) and \( y = -10 \), so \( (0, -10) \). Slope from those points: \( \frac{0 - (-10)}{4 - 0} = \frac{10}{4} = \frac{5}{2} \). Check by rearranging: \( -2y = -5x + 20 \), so \( y = \frac{5}{2}x - 10 \). Agrees. \( (4, 0) \), \( (0, -10) \), slope \( \frac{5}{2} \)

7. A concert sells adult tickets at $12 and student tickets at $7, raising $2,100. Write the constraint in standard form and find both intercepts, interpreting each.
   Show the full solution

   Let \( a \) be adult tickets and \( s \) student tickets: \[ 12a + 7s = 2100 \] Setting \( s = 0 \): \( 12a = 2100 \), so \( a = 175 \). This means 175 adult tickets and no student tickets would raise exactly $2,100. Setting \( a = 0 \): \( 7s = 2100 \), so \( s = 300 \). This means 300 student tickets and no adult tickets would do the same. Both intercepts are whole numbers here, which is fortunate; in context only whole-number points on the segment between them are actually achievable. \( 12a + 7s = 2100 \); intercepts at 175 adult-only and 300 student-only

8. Write the equation of the vertical line through \( (7, -2) \) and explain why slope-intercept form cannot express it.
   Show the full solution

   A vertical line consists of every point with the same \( x \) coordinate, so it is \( x = 7 \), which in standard form is \( 1x + 0y = 7 \). Slope-intercept form is \( y = mx + b \), which assigns a single \( y \) value to each \( x \), so it can only describe graphs that are functions. A vertical line is not a function, and its slope is undefined, so there is no value of \( m \) that could be used. Standard form has no such restriction because it allows \( B = 0 \), which is exactly the case slope-intercept form cannot reach. \( x = 7 \); slope-intercept form requires a defined slope and a function, and a vertical line has neither

9. A student finds the \( y \) intercept of \( 6x - 3y = 18 \) as \( (0, 6) \). Find the error.
   Show the full solution

   Setting \( x = 0 \) gives \( -3y = 18 \). Dividing by \( -3 \) gives \( y = -6 \), not \( +6 \). The student divided 18 by 3 and ignored the sign on the coefficient. The \( y \) intercept is \( (0, -6) \). Check in the original: \( 6(0) - 3(-6) = 0 + 18 = 18 \). Correct. Their point gives \( -3(6) = -18 \neq 18 \). \( (0, -6) \); the negative coefficient was ignored

10. Explain why the slope of \( Ax + By = C \) is \( -\frac{A}{B} \), and state when the shortcut fails.
    Show the full solution

    Start from \( Ax + By = C \) and solve for \( y \), which is what slope-intercept form requires. Subtract \( Ax \) from both sides: \[ By = -Ax + C \] Divide every term by \( B \), which requires \( B \neq 0 \): \[ y = -\frac{A}{B}x + \frac{C}{B} \] Comparing with \( y = mx + b \), the coefficient of \( x \) is \( -\frac{A}{B} \), so that is the slope, and the intercept is \( \frac{C}{B} \). The shortcut fails exactly when \( B = 0 \), because the division is not permitted. That case is a vertical line, \( Ax = C \), whose slope is undefined rather than being some number the formula failed to find. So the condition for the shortcut is the same condition as for the line being a function. It comes from solving for \( y \); it fails when \( B = 0 \), which is a vertical line with undefined slope

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. State the slope of any line parallel to \( y = 5x - 2 \).
   Show the full solution

   5

2. State the slope of any line perpendicular to \( y = 5x - 2 \).
   Show the full solution

   Flip and change sign. \( -\frac{1}{5} \)

3. State the slope perpendicular to a line of slope \( -\frac{3}{4} \).
   Show the full solution

   Flip to \( -\frac{4}{3} \), change sign to \( \frac{4}{3} \). \( \frac{4}{3} \)

4. Are \( y = 3x + 1 \) and \( y = 3x - 7 \) parallel?
   Show the full solution

   Equal slopes, different intercepts. Yes

5. What is the product of the slopes of two perpendicular lines?
   Show the full solution

   \( -1 \)

6. Are \( 2x + 3y = 6 \) and \( y = \frac{3}{2}x + 1 \) perpendicular? Justify.
   Show the full solution

   Rearrange the first: \( 3y = -2x + 6 \), so \( y = -\frac{2}{3}x + 2 \), slope \( -\frac{2}{3} \). The second has slope \( \frac{3}{2} \). Product: \( -\frac{2}{3} \times \frac{3}{2} = -\frac{6}{6} = -1 \). The product is \( -1 \), so they are perpendicular. Reading the coefficient 2 off the first equation without rearranging would have given the wrong slope entirely. Yes; the slopes are \( -\frac{2}{3} \) and \( \frac{3}{2} \), whose product is \( -1 \)

7. Write the equation of the line through \( (2, 5) \) parallel to \( y = -4x + 9 \).
   Show the full solution

   Parallel means the same slope, \( -4 \). Point-slope: \( y - 5 = -4(x - 2) \), so \( y - 5 = -4x + 8 \) and \( y = -4x + 13 \). Check: at \( x = 2 \), \( -8 + 13 = 5 \). Correct, and the slope matches while the intercept differs, so the lines are parallel and distinct. \( y = -4x + 13 \)

8. Write the equation of the line through \( (-1, 6) \) perpendicular to \( y = 2x - 3 \).
   Show the full solution

   The given slope is 2, so the perpendicular slope is \( -\frac{1}{2} \). Check: \( 2 \times -\frac{1}{2} = -1 \). Correct. Point-slope: \( y - 6 = -\frac{1}{2}(x - (-1)) = -\frac{1}{2}(x + 1) \). Distributing: \( y - 6 = -\frac{1}{2}x - \frac{1}{2} \), so \( y = -\frac{1}{2}x + \frac{11}{2} \). Check at \( x = -1 \): \( \frac{1}{2} + \frac{11}{2} = \frac{12}{2} = 6 \). Correct. \( y = -\frac{1}{2}x + \frac{11}{2} \)

9. A student says the perpendicular to a line of slope \( \frac{5}{2} \) has slope \( \frac{2}{5} \). Correct them and show the check that catches it.
   Show the full solution

   They flipped the fraction but did not change the sign. The perpendicular slope is \( -\frac{2}{5} \). The check that catches it is multiplying: their answer gives \( \frac{5}{2} \times \frac{2}{5} = 1 \), not \( -1 \). A product of \( +1 \) means the two lines are not perpendicular at all. With the correct slope, \( \frac{5}{2} \times -\frac{2}{5} = -1 \), as required. The product test is quick and unambiguous, which is why it is worth doing every time. \( -\frac{2}{5} \); their slopes multiply to \( +1 \) rather than \( -1 \)

10. Show that the points \( A(1, 2) \), \( B(5, 4) \) and \( C(3, 8) \) form a right angle at \( B \).
    Show the full solution

    A right angle at \( B \) means the two segments meeting there, \( BA \) and \( BC \), are perpendicular. Find both slopes. Slope of \( BA \), from \( B(5,4) \) to \( A(1,2) \): \( \frac{2 - 4}{1 - 5} = \frac{-2}{-4} = \frac{1}{2} \). Slope of \( BC \), from \( B(5,4) \) to \( C(3,8) \): \( \frac{8 - 4}{3 - 5} = \frac{4}{-2} = -2 \). Product: \( \frac{1}{2} \times (-2) = -1 \). Since the product is \( -1 \), the two segments are perpendicular and the angle at \( B \) is a right angle. Checking the third slope confirms the triangle is not degenerate: \( AC \) has slope \( \frac{8 - 2}{3 - 1} = 3 \), which matches neither of the others, so the three points are not collinear and a genuine right triangle is formed. The slopes at \( B \) are \( \frac{1}{2} \) and \( -2 \), whose product is \( -1 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. A job pays $18 per hour with no other payment. Write the earnings function.
   Show the full solution

   No starting value, so the intercept is 0. \( E(h) = 18h \) dollars for \( h \) hours

2. A savings account starts with $200 and gains $30 per week. Write the balance function.
   Show the full solution

   \( B(w) = 30w + 200 \) dollars after \( w \) weeks

3. A 500 page book is read at 25 pages per day. Write the pages-remaining function.
   Show the full solution

   Remaining pages decrease, so the slope is negative. \( P(d) = -25d + 500 \)

4. What feature of a model does the starting value correspond to?
   Show the full solution

   The \( y \) intercept, the output when the input is zero

5. What sign does the slope have for a quantity that is decreasing?
   Show the full solution

   Negative

6. A car's value is $24,000 new and drops $1,800 per year. Write the model and find its value after 7 years.
   Show the full solution

   Let \( y \) be years since purchase and \( V \) the value in dollars. \( V(y) = -1800y + 24000 \). At \( y = 7 \): \( -1800(7) + 24000 = -12600 + 24000 = \$11{,}400 \). The model would reach zero at \( y = 24000 \div 1800 = 13.\overline{3} \) years, after which it stops being meaningful, so the sensible domain is roughly \( 0 \leq y \leq 13 \). \( V(y) = -1800y + 24000 \); $11,400 after 7 years

7. A plant is 12 cm tall after 3 weeks and 20 cm tall after 7 weeks, growing at a constant rate. Write the model and find its height when planted.
   Show the full solution

   Points: \( (3, 12) \) and \( (7, 20) \). Slope: \( \frac{20 - 12}{7 - 3} = \frac{8}{4} = 2 \) cm per week. Point-slope with \( (3, 12) \): \( h - 12 = 2(w - 3) \), so \( h = 2w + 6 \). At planting, \( w = 0 \), giving \( h = 6 \) cm. Check the other point: \( 2(7) + 6 = 20 \). Correct. \( h(w) = 2w + 6 \); 6 cm when planted

8. A pool holds 15,000 liters and is filled at 250 liters per minute. Write the model, state the domain, and find how long filling takes.
   Show the full solution

   Let \( t \) be minutes and \( V \) the volume in liters. Starting empty: \( V(t) = 250t \). Full when \( 250t = 15000 \), so \( t = 60 \) minutes, one hour. The domain in context is \( 0 \leq t \leq 60 \) minutes, since before zero there is no filling and after 60 the pool overflows rather than continuing to follow the model. The range is \( 0 \leq V \leq 15000 \) liters. \( V(t) = 250t \), \( 0 \leq t \leq 60 \); 60 minutes to fill

9. A student models a draining tank as \( V = 40t + 600 \) and concludes the tank fills up. Diagnose the error.
   Show the full solution

   The slope is positive, which says the volume increases by 40 liters each hour. A draining tank loses water, so the volume must decrease as time increases and the slope must be negative: \( V = -40t + 600 \). The student's model is internally consistent and describes the wrong situation entirely, which is why checking the direction against one sentence of the problem is worth doing before any calculation. Their version also has no sensible domain limit, since it grows without bound, whereas the correct model empties at \( t = 15 \) hours. \( V = -40t + 600 \); the slope must be negative for a decreasing quantity

10. A taxi company charges a flag fee plus a per-mile rate. A 4 mile ride costs $14.50 and a 9 mile ride costs $28.00. Find both parameters and the cost of a 12 mile ride.
    Show the full solution

    Points: \( (4, 14.50) \) and \( (9, 28.00) \). Slope: \( \frac{28.00 - 14.50}{9 - 4} = \frac{13.50}{5} = 2.70 \) dollars per mile. Point-slope with \( (4, 14.50) \): \( C - 14.50 = 2.70(m - 4) \), so \( C - 14.50 = 2.70m - 10.80 \) and \( C = 2.70m + 3.70 \). The flag fee is the cost at zero miles, $3.70, and the per-mile rate is $2.70. Check the other point: \( 2.70(9) + 3.70 = 24.30 + 3.70 = 28.00 \). Correct. A 12 mile ride costs \( 2.70(12) + 3.70 = 32.40 + 3.70 = \$36.10 \). $3.70 flag fee and $2.70 per mile; a 12 mile ride costs $36.10

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. In \( C = 15h + 40 \) for cost in dollars and \( h \) hours, interpret the slope.
   Show the full solution

   The cost increases by $15 for each additional hour

2. In the same model, interpret the intercept.
   Show the full solution

   The cost is $40 before any hours are worked, a fixed charge

3. Define interpolation.
   Show the full solution

   Predicting a value inside the range of the data the model was built from

4. Define extrapolation.
   Show the full solution

   Predicting outside the range of the data, which is less reliable

5. What does a negative predicted volume indicate?
   Show the full solution

   That the model has been used outside its domain of validity

6. A model gives a town's population as \( P = 340t + 12400 \), \( t \) years since 2010. Interpret both parameters and predict the 2018 population.
   Show the full solution

   Slope: the population grows by 340 people per year. Intercept: the population was 12,400 in 2010, when \( t = 0 \). For 2018, \( t = 8 \): \( P = 340(8) + 12400 = 2720 + 12400 = 15{,}120 \) people. Since 2018 is likely within or near the data range, this is interpolation and is reasonably reliable. Note that people come in whole numbers, and 15,120 already is one. 340 people per year from a 2010 base of 12,400; about 15,120 in 2018

7. A model built from data for 1 to 10 years predicts a value at 60 years. Comment on the prediction.
   Show the full solution

   Sixty is six times the largest observed input, so this is extrapolation far outside the data. The arithmetic may be flawless and the prediction is still unreliable, because the model has no evidence at all that the pattern continues past 10 years. Real processes commonly change rate, level off, or reverse. The correct response is to report the computed value while stating clearly that it assumes the linear trend continued unchanged for fifty years beyond the evidence. It is extrapolation well outside the data and should be reported with the assumption stated, not as a reliable prediction

8. A shoe size model gives \( S = 0.24L - 22 \) for foot length \( L \) in cm. Explain why the intercept is not meaningful here.
   Show the full solution

   The intercept is the output when \( L = 0 \), which would be a shoe size of \( -22 \) for a foot of zero length. Neither a zero-length foot nor a negative shoe size exists, so the intercept describes nothing real. It is simply the number the line needs in order to pass through the region where the data actually lies, which for adult feet is somewhere around 22 to 30 cm. This is common in models fitted to data far from the origin, and the right interpretation is that the intercept is a fitting constant with no physical meaning rather than a starting value. An input of zero describes a foot of no length, which does not exist, so the intercept is a fitting constant rather than a real starting value

9. Two students interpret a slope of \( -2.5 \) in a model of ice thickness in cm against days. One says "it goes down by 2.5". Improve the interpretation.
   Show the full solution

   The statement is incomplete in three ways: it does not name which quantity is decreasing, it does not give the unit, and it does not say per what. A complete interpretation names both variables, the amount, the direction and both units: the ice thickness decreases by 2.5 cm for each additional day. Adding context strengthens it further: the ice is melting at a rate of 2.5 cm per day. Every interpretation question in this course is marked against that standard, and the three missing elements are where marks are lost. The ice thickness decreases by 2.5 cm for each additional day

10. A linear model of a phone battery's charge gives \( B = -12t + 100 \) percent after \( t \) hours. Find when it reaches zero, and explain why the model is likely wrong near the ends.
    Show the full solution

    Setting \( B = 0 \): \( -12t + 100 = 0 \), so \( 12t = 100 \) and \( t = 8.\overline{3} \) hours, about 8 hours 20 minutes. The domain of validity is therefore at most \( 0 \leq t \leq 8.33 \), since a charge below 0 percent or above 100 percent is impossible. The model is likely wrong near the ends for a physical reason. Battery discharge is not usually constant: many devices drain faster when the charge is high and the processor is busy, and the reported percentage often falls unevenly near empty as the voltage curve flattens. A straight line is a reasonable approximation across the middle of the range, which is where any data would have been collected, and a poorer one at the extremes. That is the general pattern with linear models: they describe the middle of the observed range well and the edges badly. Zero at about 8 hours 20 minutes; discharge is generally not at a constant rate, so the linear fit is worst at the extremes

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does the solution of a two-variable system represent graphically?
   Show the full solution

   The point where the two lines intersect

2. Is \( (3, 1) \) a solution of \( y = x - 2 \) and \( y = -2x + 7 \)?
   Show the full solution

   First: \( 3 - 2 = 1 \). Correct. Second: \( -6 + 7 = 1 \). Correct. Both hold. Yes

3. Is \( (1, 4) \) a solution of \( y = 3x + 1 \) and \( y = x + 5 \)?
   Show the full solution

   First: \( 3 + 1 = 4 \). Correct. Second: \( 1 + 5 = 6 \neq 4 \). Fails the second. No

4. How many solutions does a system of two parallel lines have?
   Show the full solution

   None

5. How many solutions if both equations describe the same line?
   Show the full solution

   Infinitely many

6. Without graphing, decide how many solutions \( y = 4x - 1 \) and \( y = -2x + 11 \) have, then find them.
   Show the full solution

   The slopes 4 and \( -2 \) differ, so the lines cross exactly once and there is one solution. Setting them equal: \( 4x - 1 = -2x + 11 \). Add \( 2x \): \( 6x - 1 = 11 \). Add 1: \( 6x = 12 \), so \( x = 2 \), and \( y = 4(2) - 1 = 7 \). Check both: \( 4(2) - 1 = 7 \) and \( -2(2) + 11 = 7 \). Correct. One solution, \( (2, 7) \)

7. Decide how many solutions \( 2x + y = 6 \) and \( 4x + 2y = 12 \) have.
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   Rearranging both: the first gives \( y = -2x + 6 \). The second gives \( 2y = -4x + 12 \), so \( y = -2x + 6 \). The two equations describe exactly the same line, since the second is just the first multiplied through by 2. Every point on that line satisfies both. Infinitely many solutions

8. Decide how many solutions \( y = \frac{1}{2}x + 3 \) and \( x - 2y = 10 \) have.
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   Rearrange the second: \( -2y = -x + 10 \), so \( y = \frac{1}{2}x - 5 \). Both lines have slope \( \frac{1}{2} \), so they are equally steep. Their intercepts are 3 and \( -5 \), which differ, so they are distinct parallel lines and never meet. No solution

9. A student solves a system by graphing and reads the intersection as \( (1.5, 2.5) \). Explain why algebraic confirmation is needed.
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   A graph is read by eye, and an intersection that falls between gridlines cannot be located exactly by looking. The true solution might be \( (1.5, 2.5) \), or it might be \( \left(\frac{14}{9}, \frac{23}{9}\right) \), which plots in almost the same place. Graphing reliably tells you how many solutions there are and roughly where, which is genuinely useful, but the exact values require substituting into both equations or solving algebraically. This limitation is the reason substitution and elimination exist. A graph gives an estimate; only substitution into both equations or algebra confirms exact values

10. Explain why a point on one line is not necessarily a solution of the system, using a specific example.
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    A single linear equation in two variables has infinitely many solutions: every point on its line satisfies it. So satisfying one equation is easy and says almost nothing. A solution to the system must satisfy every equation simultaneously, which for two crossing lines narrows the possibilities from infinitely many down to exactly one. For example, in the system \( y = x + 1 \) and \( y = 3x - 5 \), the point \( (10, 11) \) satisfies the first equation perfectly, since \( 10 + 1 = 11 \). But the second gives \( 3(10) - 5 = 25 \neq 11 \), so it fails. The actual solution is \( (3, 4) \), which satisfies both. Checking only the first equation would have accepted an endless supply of wrong answers. One equation has infinitely many solutions; only the point satisfying both qualifies, as \( (10, 11) \) shows for \( y = x + 1 \) and \( y = 3x - 5 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( y = 2x \) and \( x + y = 9 \).
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   Substitute: \( x + 2x = 9 \), so \( 3x = 9 \) and \( x = 3 \), giving \( y = 6 \). \( (3, 6) \)

2. Solve \( x = y + 4 \) and \( 2x + y = 14 \).
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   \( 2(y + 4) + y = 14 \), so \( 3y + 8 = 14 \), \( y = 2 \), and \( x = 6 \). \( (6, 2) \)

3. Solve \( y = 3x - 1 \) and \( y = x + 5 \).
   Show the full solution

   \( 3x - 1 = x + 5 \), so \( 2x = 6 \), \( x = 3 \), and \( y = 8 \). \( (3, 8) \)

4. In substitution, why must you substitute into the other equation?
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   Substituting into the same equation gives an identity and no new information

5. Solve \( y = -2x + 1 \) and \( 4x + y = 7 \).
   Show the full solution

   \( 4x + (-2x + 1) = 7 \), so \( 2x + 1 = 7 \), \( x = 3 \), and \( y = -6 + 1 = -5 \). \( (3, -5) \)

6. Solve \( 2x - y = 5 \) and \( 3x + 2y = 4 \) by substitution.
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   The easiest variable to isolate is \( y \) in the first equation, since its coefficient is \( -1 \). From \( 2x - y = 5 \), we get \( -y = -2x + 5 \), so \( y = 2x - 5 \). Substitute into the second: \( 3x + 2(2x - 5) = 4 \), giving \( 3x + 4x - 10 = 4 \), so \( 7x = 14 \) and \( x = 2 \). Back-substitute: \( y = 2(2) - 5 = -1 \). Check: \( 2(2) - (-1) = 4 + 1 = 5 \), and \( 3(2) + 2(-1) = 6 - 2 = 4 \). Both correct. \( (2, -1) \)

7. Solve \( x + 3y = 7 \) and \( 2x - y = 7 \).
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   Isolate \( x \) in the first: \( x = 7 - 3y \). Substitute: \( 2(7 - 3y) - y = 7 \), giving \( 14 - 6y - y = 7 \), so \( -7y = -7 \) and \( y = 1 \). Back-substitute: \( x = 7 - 3(1) = 4 \). Check: \( 4 + 3 = 7 \) and \( 8 - 1 = 7 \). Both correct. \( (4, 1) \)

8. Solve \( y = \frac{1}{2}x + 1 \) and \( 3x - 2y = 10 \).
   Show the full solution

   Substitute with brackets: \( 3x - 2\left(\frac{1}{2}x + 1\right) = 10 \). Distributing: \( 3x - x - 2 = 10 \), so \( 2x = 12 \) and \( x = 6 \). Back-substitute: \( y = \frac{1}{2}(6) + 1 = 4 \). Check: \( 3(6) - 2(4) = 18 - 8 = 10 \). Correct. Note that \( -2 \) times \( \frac{1}{2}x \) is \( -x \), and \( -2 \) times \( +1 \) is \( -2 \); both signs matter. \( (6, 4) \)

9. A student solves a system, finds \( x = 5 \), and writes that as the answer. Explain what is missing.
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   A system in two variables asks for values of both, so the answer is an ordered pair. Finding \( x = 5 \) locates the solution on a vertical line but does not say where on that line it sits. The remaining step is to back-substitute \( x = 5 \) into either original equation and solve for \( y \), then state the answer as \( (5, y) \) with the value found. Stopping halfway is one of the most common ways to lose marks on a system, and it is entirely avoidable by remembering that the question has two unknowns. The \( y \) value; the answer must be an ordered pair

10. Solve \( 4x + y = 10 \) and \( 8x + 2y = 20 \) by substitution, and interpret the result.
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    Isolate \( y \) in the first: \( y = 10 - 4x \). Substitute into the second: \( 8x + 2(10 - 4x) = 20 \), giving \( 8x + 20 - 8x = 20 \), so \( 20 = 20 \). The variable has vanished and what remains is true. As in lesson 2.4, that means every point satisfying the first equation also satisfies the second, so there are infinitely many solutions. The reason is visible on inspection: the second equation is exactly twice the first, so the two describe the same line. The solution set is every point on \( y = 10 - 4x \), which is best stated that way rather than as a pair of numbers. Infinitely many solutions: every point on \( y = 10 - 4x \), since the second equation is twice the first

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( x + y = 10 \) and \( x - y = 2 \).
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   The \( y \) coefficients are opposites, so add: \( 2x = 12 \), \( x = 6 \), then \( y = 4 \). \( (6, 4) \)

2. Solve \( 3x + y = 11 \) and \( 2x - y = 4 \).
   Show the full solution

   Add: \( 5x = 15 \), \( x = 3 \), then \( y = 11 - 9 = 2 \). \( (3, 2) \)

3. Solve \( 2x + 3y = 12 \) and \( 2x - y = 4 \).
   Show the full solution

   Equal \( x \) coefficients, so subtract: \( 4y = 8 \), \( y = 2 \), then \( 2x = 6 \) and \( x = 3 \). \( (3, 2) \)

4. What must you do before adding if the coefficients are 3 and 5?
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   Multiply the equations so the coefficients become opposites, such as 15 and \( -15 \)

5. When multiplying an equation by 4, how many terms are affected?
   Show the full solution

   Every term, on both sides

6. Solve \( 3x + 2y = 16 \) and \( 5x - 4y = 3 \).
   Show the full solution

   Eliminate \( y \): the coefficients 2 and \( -4 \) have least common multiple 4, so multiply the first equation by 2: \( 6x + 4y = 32 \). Now the \( y \) coefficients are \( +4 \) and \( -4 \), which are opposites, so add: \( 11x = 35 \), giving \( x = \frac{35}{11} \). That is not clean, so check the arithmetic: \( 32 + 3 = 35 \) and \( 6 + 5 = 11 \). The answer is correct even though it is not a whole number. Back-substitute: \( 3\left(\frac{35}{11}\right) + 2y = 16 \), so \( \frac{105}{11} + 2y = \frac{176}{11} \), giving \( 2y = \frac{71}{11} \) and \( y = \frac{71}{22} \). \( \left(\frac{35}{11}, \frac{71}{22}\right) \); solutions need not be whole numbers

7. Solve \( 4x + 3y = 7 \) and \( 2x + 5y = 7 \).
   Show the full solution

   Eliminate \( x \): multiply the second equation by \( -2 \), giving \( -4x - 10y = -14 \). Adding to the first: \( -7y = -7 \), so \( y = 1 \). Back-substitute into \( 4x + 3(1) = 7 \): \( 4x = 4 \), so \( x = 1 \). Check: \( 4 + 3 = 7 \) and \( 2 + 5 = 7 \). Both correct. \( (1, 1) \)

8. Solve \( 5x - 2y = 4 \) and \( 3x + 4y = 18 \).
   Show the full solution

   Eliminate \( y \): the coefficients \( -2 \) and 4 have least common multiple 4, so multiply the first by 2: \( 10x - 4y = 8 \). Adding to the second: \( 13x = 26 \), so \( x = 2 \). Back-substitute: \( 5(2) - 2y = 4 \), so \( -2y = -6 \) and \( y = 3 \). Check: \( 10 - 6 = 4 \) and \( 6 + 12 = 18 \). Both correct. \( (2, 3) \)

9. A student multiplies \( 2x + 3y = 7 \) by 4 and writes \( 8x + 12y = 7 \). Find the error.
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   They multiplied the left side but not the right. The multiplication property of equality requires both sides to be multiplied, so the correct result is \( 8x + 12y = 28 \). Their version describes a completely different line: checking with \( (2, 1) \), which satisfies the original since \( 4 + 3 = 7 \), their equation gives \( 16 + 12 = 28 \neq 7 \). Multiplying an equation by a constant preserves its solutions only if every term is multiplied. \( 8x + 12y = 28 \); the right-hand side was not multiplied

10. Solve \( 6x - 4y = 10 \) and \( -3x + 2y = -5 \), and interpret the result.
    Show the full solution

    Eliminate \( x \) by multiplying the second equation by 2: \( -6x + 4y = -10 \). Adding to the first: \( 0 = 0 \). Both variables have vanished and the statement left is true, so there are infinitely many solutions. The reason is that the second equation is exactly the first multiplied by \( -\frac{1}{2} \), so the two describe the same line. Rearranging either to slope-intercept form gives \( y = \frac{3}{2}x - \frac{5}{2} \), and every point on that line satisfies both equations. Contrast this with what a false statement such as \( 0 = 7 \) would have meant: parallel distinct lines and no solution. The test is the same as in lesson 2.4, applied to two equations at once. Infinitely many: every point on \( y = \frac{3}{2}x - \frac{5}{2} \), since the equations describe the same line

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Which method suits \( y = 4x + 1 \) and \( 3x + y = 15 \)?
   Show the full solution

   Substitution, since \( y \) is already isolated

2. Which method suits \( 5x + 2y = 9 \) and \( 5x - 3y = -1 \)?
   Show the full solution

   Elimination, since the \( x \) coefficients are already equal

3. What does ending with \( 0 = 5 \) mean?
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   No solution; the lines are parallel and distinct

4. What does ending with \( 0 = 0 \) mean?
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   Infinitely many solutions; the equations describe the same line

5. How can you spot the infinitely-many case by inspection?
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   One equation is a constant multiple of the other

6. Solve \( 3x - y = 8 \) and \( 6x - 2y = 16 \), choosing a method and justifying it.
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   Inspection first: the second equation is exactly twice the first, since \( 2(3x - y) = 6x - 2y \) and \( 2(8) = 16 \). That signals the infinitely-many case before any method is applied. Confirming by elimination: multiply the first by \( -2 \) to get \( -6x + 2y = -16 \), then add to the second: \( 0 = 0 \), a true statement. The solution set is every point on the line, which rearranged is \( y = 3x - 8 \). Infinitely many: every point on \( y = 3x - 8 \)

7. Solve \( x = 2y + 1 \) and \( 3x - 6y = 5 \).
   Show the full solution

   Substitution is natural since \( x \) is isolated: \( 3(2y + 1) - 6y = 5 \), giving \( 6y + 3 - 6y = 5 \), so \( 3 = 5 \). A false statement, so there is no solution. Confirming with slopes: the first rearranges to \( y = \frac{1}{2}x - \frac{1}{2} \) and the second to \( y = \frac{1}{2}x - \frac{5}{6} \). Equal slopes, different intercepts: distinct parallel lines. No solution

8. Solve \( 7x + 4y = 1 \) and \( 7x - 2y = 13 \).
   Show the full solution

   Elimination, since the \( x \) coefficients are equal. Subtract the second from the first: \( 6y = -12 \), so \( y = -2 \). Back-substitute: \( 7x + 4(-2) = 1 \), so \( 7x = 9 \) and \( x = \frac{9}{7} \). Check in the second: \( 7\left(\frac{9}{7}\right) - 2(-2) = 9 + 4 = 13 \). Correct. \( \left(\frac{9}{7}, -2\right) \)

9. A student solves a system, gets \( 4 = 4 \), and writes "no solution". Correct them.
   Show the full solution

   They have the two cases reversed. Ending with a true statement means the two equations impose the same requirement, so every point satisfying one satisfies the other and there are infinitely many solutions. Ending with a false statement, such as \( 4 = 9 \), would mean the requirements contradict each other and nothing works, which is the no-solution case. The reading is straightforward once stated in words: a true leftover says "always", a false leftover says "never". Infinitely many solutions; a true leftover statement means the equations agree everywhere

10. Find the value of \( k \) making \( 2x + 6y = 10 \) and \( x + 3y = k \) have infinitely many solutions, and state what happens otherwise.
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    Infinitely many solutions requires the two equations to describe the same line, which means one must be a constant multiple of the other. Comparing the variable terms, \( 2x + 6y \) is exactly twice \( x + 3y \), so the multiplier is 2. For the equations to match, the constants must be in the same ratio: \( 10 = 2k \), giving \( k = 5 \). Checking: with \( k = 5 \), the second equation doubled is \( 2x + 6y = 10 \), which is the first exactly. For any other value of \( k \), the variable terms are still proportional but the constants are not, so the lines have the same slope and different intercepts: distinct parallel lines with no solution. There is no value of \( k \) giving exactly one solution, because the slopes are forced to be equal by the variable coefficients. \( k = 5 \); every other value gives no solution, and none gives exactly one

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Two numbers sum to 20 and differ by 4. Write the system.
   Show the full solution

   \( x + y = 20 \) and \( x - y = 4 \)

2. Solve that system.
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   Add: \( 2x = 24 \), \( x = 12 \), then \( y = 8 \). 12 and 8

3. How many independent constraints does a two-variable system need?
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   Two

4. If \( x \) items cost $5 each and \( y \) items cost $9 each, write the total cost.
   Show the full solution

   \( 5x + 9y \)

5. Why must the two equations carry different information?
   Show the full solution

   Two statements of the same constraint give the same line and infinitely many solutions

6. A farm has chickens and cows totaling 48 animals with 134 legs. How many of each?
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   Let \( c \) be chickens and \( w \) cows. Chickens have 2 legs, cows have 4. Count: \( c + w = 48 \). Legs: \( 2c + 4w = 134 \). Substitute \( c = 48 - w \): \( 2(48 - w) + 4w = 134 \), giving \( 96 - 2w + 4w = 134 \), so \( 2w = 38 \) and \( w = 19 \). Then \( c = 29 \). Check: \( 29 + 19 = 48 \), and \( 2(29) + 4(19) = 58 + 76 = 134 \). Correct, and both are whole positive numbers as animals must be. 29 chickens and 19 cows

7. A boat travels 60 km downstream in 3 hours and 60 km upstream in 5 hours. Find the boat's speed in still water and the current's speed.
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   Let \( b \) be the boat's speed in still water and \( r \) the current's speed, both in km/h. Downstream the current helps, so the effective speed is \( b + r \); upstream it hinders, so it is \( b - r \). Using distance divided by time: Downstream: \( b + r = 60 \div 3 = 20 \). Upstream: \( b - r = 60 \div 5 = 12 \). Adding: \( 2b = 32 \), so \( b = 16 \). Then \( r = 20 - 16 = 4 \). Check: downstream \( 16 + 4 = 20 \) km/h covers 60 km in 3 hours; upstream \( 16 - 4 = 12 \) km/h covers 60 km in 5 hours. Correct. Boat 16 km/h, current 4 km/h

8. A chemist mixes a 20 percent solution with a 50 percent solution to make 60 mL of a 30 percent solution. How much of each?
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   Let \( x \) be mL of the 20 percent solution and \( y \) mL of the 50 percent. Volume: \( x + y = 60 \). Amount of the active substance: the 20 percent solution contributes \( 0.20x \) mL, the 50 percent contributes \( 0.50y \) mL, and the result must contain \( 0.30(60) = 18 \) mL: \( 0.20x + 0.50y = 18 \). Substitute \( y = 60 - x \): \( 0.20x + 0.50(60 - x) = 18 \), giving \( 0.20x + 30 - 0.50x = 18 \), so \( -0.30x = -12 \) and \( x = 40 \). Then \( y = 20 \). Check: \( 0.20(40) + 0.50(20) = 8 + 10 = 18 \) mL in 60 mL, which is 30 percent. Correct. 40 mL of the 20 percent and 20 mL of the 50 percent

9. A student sets up "the total is 45 items" and "there are 45 things altogether" as two equations. Explain what is wrong.
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   Both sentences say the same thing, so both produce the equation \( x + y = 45 \). Two identical equations describe one line, so the system has infinitely many solutions and determines nothing. A system needs two independent constraints, meaning two facts that could vary separately. The second fact must measure something different: a total cost, a total weight, a difference, or a ratio. Without it, the problem as stated does not have a unique answer, and no amount of algebra will produce one. Both equations are the same constraint, so the system is not independent and has infinitely many solutions

10. A coin jar holds only nickels and dimes, 40 coins worth $3.10. Find how many of each, and explain what makes the answer unique.
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    Let \( n \) be the number of nickels and \( d \) the number of dimes. Count: \( n + d = 40 \). Value, in cents to avoid decimals: \( 5n + 10d = 310 \). Substitute \( n = 40 - d \): \( 5(40 - d) + 10d = 310 \), giving \( 200 - 5d + 10d = 310 \), so \( 5d = 110 \) and \( d = 22 \). Then \( n = 18 \). Check: \( 18 + 22 = 40 \) coins, and \( 5(18) + 10(22) = 90 + 220 = 310 \) cents, which is $3.10. Correct, and both counts are whole and positive. The answer is unique because the two constraints are genuinely independent: one counts coins and the other totals value, and the two coin types have different values. Had both coins been worth the same, the value equation would have been a multiple of the count equation and the system would not have determined a unique split. 18 nickels and 22 dimes; the two constraints measure different quantities, so they are independent

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Should \( y \lt 3x + 1 \) be graphed with a solid or dashed line?
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   Strict inequality excludes the boundary. Dashed

2. Should \( y \geq x - 4 \) be graphed with a solid or dashed line?
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   Solid

3. Is \( (0, 0) \) a solution of \( x + y \leq 5 \)?
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   \( 0 \leq 5 \) is true. Yes

4. Is \( (4, 3) \) a solution of \( y \gt 2x \)?
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   \( 3 \gt 8 \) is false. No

5. What does the shaded region of an inequality represent?
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   Every point whose coordinates satisfy the inequality

6. Describe how to graph \( 3x - y \gt 6 \), including the test.
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   Boundary: \( 3x - y = 6 \). Intercepts: setting \( y = 0 \) gives \( x = 2 \), so \( (2, 0) \); setting \( x = 0 \) gives \( -y = 6 \), so \( y = -6 \) and \( (0, -6) \). The inequality is strict, so the line is dashed. Test the origin, which is not on the line: \( 3(0) - 0 = 0 \), and \( 0 \gt 6 \) is false. So the origin's side is not shaded; shade the other side, below and to the right of the line. Verify with \( (4, 0) \): \( 12 - 0 = 12 \gt 6 \), true, confirming that side. Dashed line through \( (2, 0) \) and \( (0, -6) \), shading the side away from the origin

7. Is \( (3, -2) \) a solution of \( 4x + 5y \leq 1 \)?
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   \( 4(3) + 5(-2) = 12 - 10 = 2 \), and \( 2 \leq 1 \) is false. No

8. Explain why the origin cannot always be used as a test point, and what to use instead.
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   A test point works by being clearly on one side of the boundary, so that whether it satisfies the inequality identifies which side to shade. If the boundary line passes through the origin, the origin is on the line rather than on either side, so substituting it gives no information about the sides. Any line of the form \( y = mx \) or \( ax + by = 0 \) has this property. In that case choose any other point that is clearly off the line, such as \( (1, 0) \) or \( (0, 1) \), and test it instead. The method is unchanged; only the convenient choice of point differs. If the line passes through the origin, use another point such as \( (1, 0) \) or \( (0, 1) \)

9. A student graphs \( 2x + y \geq 8 \) and shades the side containing the origin. Check and correct.
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   Test the origin in the original: \( 2(0) + 0 = 0 \), and \( 0 \geq 8 \) is false. So the origin does not satisfy the inequality and its side must not be shaded. The correct shading is the other side, away from the origin. The student probably reasoned from the \( \geq \) symbol, expecting to shade "above", without noticing that the inequality is in standard form where that rule does not apply directly. Testing a point makes the rule unnecessary and cannot be misapplied. Verifying the correction: \( (5, 0) \) gives \( 10 \geq 8 \), true, and it lies on the far side from the origin. Shade the side away from the origin; the origin gives \( 0 \geq 8 \), which is false

10. A student has $60 to spend on notebooks at $4 and pens at $2. Write an inequality, describe its graph, and explain which part of the region is actually meaningful.
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    Let \( n \) be the number of notebooks and \( p \) the number of pens. Spending must not exceed $60: \[ 4n + 2p \leq 60 \] Boundary: \( 4n + 2p = 60 \), with intercepts at \( n = 15 \) when \( p = 0 \), and \( p = 30 \) when \( n = 0 \). The inequality includes equality, so the line is solid. Testing the origin: \( 0 \leq 60 \), true, so the region containing the origin is shaded, which is everything below the line. The meaningful part is smaller than the shaded half-plane. Both quantities are counts, so they must be whole numbers and cannot be negative: \( n \geq 0 \) and \( p \geq 0 \). The realistic solution set is therefore the lattice points with whole-number coordinates inside the triangle bounded by the two axes and the line. Points such as \( (2.5, 7) \) satisfy the inequality but do not correspond to a possible purchase. \( 4n + 2p \leq 60 \); only the whole-number points in the first quadrant part of the region are meaningful

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What region represents the solution of a system of inequalities?
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   The overlap, or intersection, of all the shaded regions

2. What do \( x \geq 0 \) and \( y \geq 0 \) restrict the graph to?
   Show the full solution

   The first quadrant, including both axes

3. Is \( (1, 1) \) a solution of \( x + y \leq 5 \) and \( y \geq x \)?
   Show the full solution

   \( 2 \leq 5 \) is true, and \( 1 \geq 1 \) is true. Yes

4. Is \( (4, 1) \) a solution of \( x + y \leq 5 \) and \( y \geq x \)?
   Show the full solution

   \( 5 \leq 5 \) is true, but \( 1 \geq 4 \) is false. It must satisfy both. No

5. What is the feasible region?
   Show the full solution

   The set of points satisfying every constraint at once

6. Find the corners of the region given by \( x + y \leq 6 \), \( x \geq 0 \), \( y \geq 0 \).
   Show the full solution

   The nonnegativity constraints confine the region to the first quadrant. The line \( x + y = 6 \) has intercepts \( (6, 0) \) and \( (0, 6) \), and testing the origin gives \( 0 \leq 6 \), true, so the region is the triangle below that line. Its corners are where the boundaries meet: the two axes meet at \( (0, 0) \), the line meets the \( x \) axis at \( (6, 0) \), and the line meets the \( y \) axis at \( (0, 6) \). \( (0, 0) \), \( (6, 0) \) and \( (0, 6) \)

7. Find the corner where \( 3x + y = 12 \) meets \( x + y = 6 \).
   Show the full solution

   The \( y \) coefficients are equal, so subtract the second from the first: \( 2x = 6 \), giving \( x = 3 \). Back-substitute into \( x + y = 6 \): \( y = 3 \). Check in both: \( 3(3) + 3 = 12 \), and \( 3 + 3 = 6 \). Correct. \( (3, 3) \)

8. Explain what it means if two shaded regions do not overlap at all.
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   A solution of the system must satisfy every inequality simultaneously, which means it must lie in every shaded region. If the regions have no points in common, then no point satisfies all the constraints and the system has no solution. In a real situation this is informative rather than a failure: it means the requirements are mutually incompatible, and the honest conclusion is that the problem as stated cannot be met. For example, requiring \( x + y \geq 10 \) and \( x + y \leq 4 \) at once is impossible, since no sum can be both at least 10 and at most 4. The system has no solution; the constraints are incompatible

9. A student shades every region and calls the whole shaded area the solution. Correct them.
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   They have found the union rather than the intersection. A point in the union satisfies at least one inequality, which is not what a system requires. A system asks for points satisfying all of them, so the solution is only the part where every shaded region overlaps, which is generally a much smaller area. The test that settles it is to take a point in a singly-shaded part and check it against every inequality: it will fail at least one, proving it is not a solution. The solution is only the overlap, where every inequality is satisfied, not the whole shaded area

10. A workshop makes tables and chairs. Each table needs 4 hours and each chair 2 hours, with 40 hours available. Wood limits them to 15 items total. Write the system and find the corners.
    Show the full solution

    Let \( t \) be tables and \( c \) chairs. Time: \( 4t + 2c \leq 40 \). Items: \( t + c \leq 15 \). Nonnegativity: \( t \geq 0 \), \( c \geq 0 \). Corners on the axes: the origin \( (0, 0) \). On the \( t \) axis, time allows \( t = 10 \) and items allow \( t = 15 \), so the binding one is \( (10, 0) \). On the \( c \) axis, time allows \( c = 20 \) and items allow \( c = 15 \), so the binding one is \( (0, 15) \). The interior corner solves \( 4t + 2c = 40 \) and \( t + c = 15 \). From the second, \( c = 15 - t \). Substituting: \( 4t + 2(15 - t) = 40 \), giving \( 4t + 30 - 2t = 40 \), so \( 2t = 10 \) and \( t = 5 \), then \( c = 10 \). Check \( (5, 10) \): time \( 20 + 20 = 40 \leq 40 \), items \( 5 + 10 = 15 \leq 15 \), both nonnegative. It satisfies all four and sits on both boundaries. \( 4t + 2c \leq 40 \), \( t + c \leq 15 \), \( t, c \geq 0 \); corners \( (0,0) \), \( (10,0) \), \( (5,10) \), \( (0,15) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the common difference of 5, 9, 13, 17, …
   Show the full solution

   \( d = 4 \)

2. Find the common difference of 20, 17, 14, 11, …
   Show the full solution

   \( 17 - 20 = -3 \). \( d = -3 \)

3. Write the explicit rule for 5, 9, 13, …
   Show the full solution

   \( a_n = 5 + 4(n-1) = 4n + 1 \). Check: \( 4(1) + 1 = 5 \). \( a_n = 4n + 1 \)

4. Is 2, 4, 8, 16 arithmetic?
   Show the full solution

   Differences are 2, 4, 8, which are not constant. No

5. Why does the explicit rule use \( n - 1 \)?
   Show the full solution

   The first term has had no differences added yet, so reaching term \( n \) takes \( n - 1 \) steps

6. For 100, 93, 86, …, find the 12th term.
   Show the full solution

   The differences are \( 93 - 100 = -7 \) and \( 86 - 93 = -7 \), so \( d = -7 \) and \( a_1 = 100 \). Explicit rule: \( a_n = 100 - 7(n-1) = 107 - 7n \). Checking at \( n = 1 \): \( 107 - 7 = 100 \). Correct. \( a_{12} = 107 - 7(12) = 107 - 84 = 23 \). 23

7. In that same sequence, which term equals 2?
   Show the full solution

   Set \( 107 - 7n = 2 \), so \( -7n = -105 \) and \( n = 15 \). Check: \( 107 - 105 = 2 \). Since 15 is a positive whole number, 2 is the 15th term. The 15th term

8. An arithmetic sequence has \( a_3 = 14 \) and \( a_7 = 30 \). Find the explicit rule.
   Show the full solution

   Going from term 3 to term 7 is 4 steps, and the value rises by \( 30 - 14 = 16 \). So \( 4d = 16 \) and \( d = 4 \). Working back to the first term: \( a_1 = a_3 - 2d = 14 - 8 = 6 \). So \( a_n = 6 + 4(n-1) = 4n + 2 \). Check both given terms: \( a_3 = 4(3) + 2 = 14 \), and \( a_7 = 4(7) + 2 = 30 \). Both correct. \( a_n = 4n + 2 \)

9. Convert \( a_1 = 9 \), \( a_n = a_{n-1} - 5 \) to explicit form, and explain the advantage.
   Show the full solution

   The recursive rule says the first term is 9 and each term is 5 less than the one before, so \( d = -5 \). Explicit: \( a_n = 9 - 5(n-1) = 14 - 5n \). Check at \( n = 1 \): \( 14 - 5 = 9 \). Correct. The advantage is direct access. The recursive rule requires computing every earlier term first, so finding \( a_{200} \) would mean 199 subtractions. The explicit rule gives \( a_{200} = 14 - 1000 = -986 \) in one step. Recursive rules describe the pattern more naturally; explicit rules compute far terms efficiently, which is why translating between them is worth the practice. \( a_n = 14 - 5n \); it gives any term directly rather than requiring all the earlier ones

10. A theater has 18 seats in the first row and 2 more in each row after. Find the seats in row 25 and the total in the first 5 rows, then say why the model might fail.
    Show the full solution

    The row counts form an arithmetic sequence with \( a_1 = 18 \) and \( d = 2 \), so \( a_n = 18 + 2(n-1) = 2n + 16 \). Checking: \( 2(1) + 16 = 18 \). Correct. Row 25: \( a_{25} = 2(25) + 16 = 66 \) seats. First five rows: 18, 20, 22, 24, 26, which sum to \( 18 + 20 + 22 + 24 + 26 = 110 \) seats. The model assumes the pattern continues without limit. In a real theater the room has a fixed width, so at some row the wall stops the growth and the counts level off or the rows end. A sequence can only be trusted within the range where its assumption holds, which is the same caution that applies to every model in this course. 66 seats in row 25, 110 seats in the first five rows; the model assumes unlimited width

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the common ratio of 5, 15, 45, 135, …
   Show the full solution

   \( 15 \div 5 = 3 \). \( r = 3 \)

2. Find the common ratio of 80, 40, 20, 10, …
   Show the full solution

   \( r = \frac{1}{2} \)

3. Write the explicit rule for 5, 15, 45, …
   Show the full solution

   \( a_n = 5 \cdot 3^{\,n-1} \)

4. Is 3, 7, 11, 15 arithmetic or geometric?
   Show the full solution

   Differences are all 4; ratios are not constant. Arithmetic

5. Evaluate \( a_n = 4 \cdot 3^{\,n-1} \) at \( n = 3 \).
   Show the full solution

   \( 4 \cdot 3^2 = 4 \cdot 9 = 36 \). The exponent applies only to the 3. 36

6. Classify 2, 6, 12, 20, 30, … and justify.
   Show the full solution

   Differences: 4, 6, 8, 10. Not constant, so not arithmetic. Ratios: 3, 2, \( \frac{5}{3} \), 1.5. Not constant, so not geometric. It is neither. The differences themselves increase by 2 each time, which makes it a quadratic pattern of the kind studied in unit 8. Not every sequence falls into one of the two families in this unit, and checking both tests before deciding is the point of the question. Neither; both tests fail

7. A geometric sequence has \( a_1 = 5 \) and \( a_4 = 40 \). Find \( r \).
   Show the full solution

   Using the explicit rule: \( a_4 = 5 r^3 = 40 \), so \( r^3 = 8 \) and \( r = 2 \). Check by listing: 5, 10, 20, 40. The fourth term is 40. Correct. \( r = 2 \)

8. Find the 6th term of 1, \( -3 \), 9, \( -27 \), …
   Show the full solution

   Ratios: \( -3 \div 1 = -3 \), \( 9 \div (-3) = -3 \), \( -27 \div 9 = -3 \). Constant, so \( r = -3 \). \( a_6 = 1 \cdot (-3)^5 = -243 \), since an odd power of a negative number is negative. Check by listing: 1, \( -3 \), 9, \( -27 \), 81, \( -243 \). Correct. A negative ratio makes the signs alternate, with odd-numbered terms positive here and even-numbered terms negative. \( -243 \)

9. Explain why an arithmetic sequence with a positive difference is eventually overtaken by a geometric sequence with \( r \gt 1 \), even if the arithmetic one starts far ahead.
   Show the full solution

   An arithmetic sequence adds the same fixed amount at every step, so the amount it gains per step never changes. A geometric sequence with \( r \gt 1 \) multiplies, so the amount it gains per step is a fixed fraction of a number that is itself growing. Its increments therefore grow without bound while the arithmetic increments stay constant, so it must eventually catch up and pass no matter how large a head start it gives away. For example, compare \( a_n = 1000 + 10n \) with \( b_n = 2^n \). At \( n = 5 \) the first is 1050 and the second is 32, so the arithmetic one is far ahead. At \( n = 11 \) the first is 1110 and the second is 2048, and the geometric one has passed it permanently. Geometric increments grow proportionally while arithmetic increments stay fixed, so the geometric sequence always overtakes eventually

10. A ball dropped from 200 cm rebounds to 60 percent of its previous height. Find the height after the fourth bounce and explain why the sequence never reaches zero.
    Show the full solution

    Each bounce multiplies the height by 0.6, so the heights form a geometric sequence with \( r = 0.6 \). Taking the drop height 200 cm as the starting value, the height after \( n \) bounces is \( 200(0.6)^n \). After four bounces: \( 200(0.6)^4 \). Computing \( 0.6^2 = 0.36 \) and \( 0.6^4 = 0.36^2 = 0.1296 \), so the height is \( 200 \times 0.1296 = 25.92 \) cm. Checking bounce by bounce: 120, 72, 43.2, 25.92 cm. The two agree. Mathematically the sequence never reaches zero because multiplying a positive number by 0.6 always leaves a positive number, however small. The terms approach zero without attaining it. Physically the ball does stop, because friction and the finite size of the ball take over once the heights become tiny, so the model is a good description for the first several bounces and not for the hundredth. 25.92 cm; a positive number times 0.6 stays positive, so the terms approach zero without reaching it

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( x^4 \cdot x^6 \).
   Show the full solution

   \( x^{10} \)

2. Simplify \( \dfrac{y^9}{y^4} \).
   Show the full solution

   \( y^5 \)

3. Simplify \( (m^3)^5 \).
   Show the full solution

   \( m^{15} \)

4. Simplify \( (3a)^2 \).
   Show the full solution

   Both factors are squared. \( 9a^2 \)

5. Can \( x^3 \cdot y^5 \) be simplified?
   Show the full solution

   No; the bases differ, so no factors combine

6. Simplify \( (4x^2 y^3)^3 \).
   Show the full solution

   Every factor inside the bracket is cubed: \( 4^3 = 64 \), \( (x^2)^3 = x^6 \), and \( (y^3)^3 = y^9 \). \( 64x^6 y^9 \)

7. Simplify \( \dfrac{20a^5 b^3}{5a^2 b^7} \).
   Show the full solution

   Separate the parts: \( \dfrac{20}{5} = 4 \); \( \dfrac{a^5}{a^2} = a^3 \); \( \dfrac{b^3}{b^7} = \dfrac{1}{b^4} \). \( \dfrac{4a^3}{b^4} \)

8. Simplify \( \dfrac{(2x^3)^4}{8x^5} \).
   Show the full solution

   Handle the bracket first: \( (2x^3)^4 = 2^4 (x^3)^4 = 16x^{12} \). Then divide: \( \dfrac{16x^{12}}{8x^5} = \dfrac{16}{8} \cdot x^{12-5} = 2x^7 \). Check at \( x = 1 \): the original is \( \dfrac{2^4}{8} = 2 \), and the simplified form is 2. They agree. \( 2x^7 \)

9. A student writes \( x^3 \cdot x^4 = x^{12} \). Explain the error from the definition.
   Show the full solution

   They applied the power rule where the product rule belongs. Writing out the definition settles it: \( x^3 \cdot x^4 = (xxx)(xxxx) \), which is \( x \) used seven times, so \( x^7 \). Nothing is being grouped, so nothing multiplies; the factors from the two powers simply accumulate, which adds the counts. The multiplying rule belongs to \( (x^3)^4 \), which means four copies of \( x^3 \) and therefore twelve factors. A numerical check confirms the distinction: at \( x = 2 \), \( 2^3 \cdot 2^4 = 8 \cdot 16 = 128 = 2^7 \), while \( (2^3)^4 = 8^4 = 4096 = 2^{12} \). \( x^7 \); multiplying powers adds exponents, and only raising a power to a power multiplies them

10. Simplify \( \left(\dfrac{3x^2}{y}\right)^3 \cdot \dfrac{y^4}{9x^3} \) and check numerically.
    Show the full solution

    Apply the power of a quotient first, cubing every factor: \( \left(\dfrac{3x^2}{y}\right)^3 = \dfrac{3^3 (x^2)^3}{y^3} = \dfrac{27x^6}{y^3} \). Multiply: \( \dfrac{27x^6}{y^3} \cdot \dfrac{y^4}{9x^3} = \dfrac{27x^6 y^4}{9x^3 y^3} \). Simplify each part: \( \dfrac{27}{9} = 3 \); \( \dfrac{x^6}{x^3} = x^3 \); \( \dfrac{y^4}{y^3} = y \). The result is \( 3x^3 y \). Check at \( x = 1 \), \( y = 2 \): the original is \( \left(\dfrac{3}{2}\right)^3 \cdot \dfrac{16}{9} = \dfrac{27}{8} \cdot \dfrac{16}{9} = \dfrac{432}{72} = 6 \), and the simplified form gives \( 3(1)(2) = 6 \). They agree. \( 3x^3 y \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Evaluate \( 12^0 \).
   Show the full solution

   1

2. Evaluate \( 3^{-2} \).
   Show the full solution

   \( \dfrac{1}{3^2} = \dfrac{1}{9} \). \( \frac{1}{9} \)

3. Evaluate \( 25^{1/2} \).
   Show the full solution

   5

4. Evaluate \( 27^{1/3} \).
   Show the full solution

   3

5. Write 0.00045 in scientific notation.
   Show the full solution

   The decimal point moves four places right to reach 4.5. \( 4.5 \times 10^{-4} \)

6. Evaluate \( 27^{2/3} \) and \( 81^{3/4} \).
   Show the full solution

   For the first, take the cube root first: \( 27^{1/3} = 3 \), then \( 3^2 = 9 \). For the second, take the fourth root first: \( 81^{1/4} = 3 \), since \( 3^4 = 81 \), then \( 3^3 = 27 \). Both check by the other order: \( 27^2 = 729 \) and \( 729^{1/3} = 9 \); and \( 81^3 = 531{,}441 \) with \( 531{,}441^{1/4} = 27 \), which is correct but far more work. 9 and 27

7. Simplify \( \left(\dfrac{2}{3}\right)^{-2} \).
   Show the full solution

   A negative exponent means the reciprocal, so \( \left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4} \). Checking the long way: \( \left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9} \), and the reciprocal of \( \dfrac{4}{9} \) is \( \dfrac{9}{4} \). The two agree. \( \frac{9}{4} \)

8. Compute \( \dfrac{4.2 \times 10^5}{7 \times 10^2} \) in scientific notation.
   Show the full solution

   Separate the parts: \( \dfrac{4.2}{7} = 0.6 \), and \( \dfrac{10^5}{10^2} = 10^3 \). That gives \( 0.6 \times 10^3 \), which is not in proper form because 0.6 is less than 1. Renormalize by writing \( 0.6 = 6 \times 10^{-1} \), so the value is \( 6 \times 10^{-1} \times 10^3 = 6 \times 10^2 \). Check in ordinary notation: \( 420{,}000 \div 700 = 600 \), and \( 6 \times 10^2 = 600 \). Correct. \( 6 \times 10^2 \)

9. Explain from the quotient rule why \( x^0 = 1 \), and why \( x = 0 \) is excluded.
   Show the full solution

   Take any nonzero \( x \) and any exponent \( n \). The quotient rule gives \( \dfrac{x^n}{x^n} = x^{n-n} = x^0 \). But a nonzero quantity divided by itself is 1. Since both computations describe the same expression, \( x^0 \) must equal 1. The definition is not a convention chosen for convenience; it is the only value that keeps the quotient rule consistent. The base zero is excluded because the argument relies on dividing by \( x^n \), and if \( x = 0 \) then \( x^n = 0 \) and the division is undefined. So \( 0^0 \) is not assigned a value in this course, and the rule is always stated for nonzero bases. The quotient rule forces \( x^0 = 1 \); zero is excluded because the argument divides by \( x^n \)

10. Simplify \( \dfrac{6x^{-2} y^3}{3x^4 y^{-1}} \) with only positive exponents, and check.
    Show the full solution

    Separate the parts. Numeric: \( \dfrac{6}{3} = 2 \). For \( x \): \( \dfrac{x^{-2}}{x^4} = x^{-2-4} = x^{-6} = \dfrac{1}{x^6} \). For \( y \): \( \dfrac{y^3}{y^{-1}} = y^{3-(-1)} = y^4 \). Subtracting a negative exponent adds, which is the step most often mishandled. Combining: \( \dfrac{2y^4}{x^6} \). Check at \( x = 2 \), \( y = 1 \). Original: \( \dfrac{6 \cdot \frac{1}{4} \cdot 1}{3 \cdot 16 \cdot 1} = \dfrac{1.5}{48} = 0.03125 \). Simplified: \( \dfrac{2 \cdot 1}{64} = 0.03125 \). They agree. \( \dfrac{2y^4}{x^6} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the growth factor for a 25 percent increase?
   Show the full solution

   \( 1 + 0.25 \). 1.25

2. What is the decay factor for a 40 percent decrease?
   Show the full solution

   \( 1 - 0.40 \). 0.60

3. In \( y = 300(1.08)^x \), what is the initial value?
   Show the full solution

   300

4. Does \( y = 80(0.9)^x \) grow or decay?
   Show the full solution

   The factor is between 0 and 1. Decay, by 10 percent each step

5. Write a model for 200 growing 5 percent per year.
   Show the full solution

   \( y = 200(1.05)^t \)

6. A car worth $24,000 loses 18 percent of its value each year. Find its value after 4 years.
   Show the full solution

   Losing 18 percent means keeping 82 percent, so the factor is \( 1 - 0.18 = 0.82 \) and the model is \( V = 24000(0.82)^t \). Computing \( 0.82^2 = 0.6724 \) and \( 0.82^4 = 0.6724^2 \approx 0.452122 \). \( V = 24000 \times 0.452122 \approx 10850.93 \). Check year by year: 19,680, then 16,137.60, then 13,232.83, then 10,850.92, agreeing to within rounding. About $10,851

7. A population of 1,200 grows 3 percent per year. Find it after 20 years.
   Show the full solution

   Model: \( P = 1200(1.03)^t \). \( 1.03^{20} \approx 1.806111 \), so \( P \approx 1200 \times 1.806111 \approx 2167.3 \). Since a population is a count, report it as about 2,167 people. Note that the population has not grown by \( 20 \times 3 = 60 \) percent, which would give 1,920. It has grown by about 80.6 percent, and the extra comes from compounding. About 2,167

8. $2,000 is invested at 4 percent compounded quarterly. Find the balance after 6 years.
   Show the full solution

   Quarterly means \( n = 4 \), so each quarter the factor is \( 1 + \frac{0.04}{4} = 1.01 \), and there are \( 4 \times 6 = 24 \) quarters. \( A = 2000(1.01)^{24} \). \( 1.01^{24} \approx 1.269735 \), so \( A \approx 2000 \times 1.269735 \approx 2539.47 \). About $2,539.47

9. A student models 8 percent annual growth of $100 as \( y = 100(0.08)^t \). Explain what that equation actually describes.
   Show the full solution

   They used the rate as the factor instead of converting it. Multiplying by 0.08 each year keeps only 8 percent of the previous value, which is a loss of 92 percent per year. After one year their model gives $8 rather than $108, which is obviously wrong from the story alone. The correct factor is \( 1 + 0.08 = 1.08 \), giving \( y = 100(1.08)^t \) and a first year value of $108. The check that catches this error in a few seconds is to substitute \( t = 1 \) and ask whether the answer matches what the situation says should happen after one step. It describes a 92 percent annual loss; the correct model is \( y = 100(1.08)^t \)

10. A medication leaves the body at 30 percent per hour from a 60 mg dose. Find the amount after 5 hours, and explain why the amount never reaches zero in the model.
    Show the full solution

    Losing 30 percent per hour means retaining 70 percent, so the factor is 0.70 and the model is \( A = 60(0.7)^t \) with \( t \) in hours. Computing \( 0.7^2 = 0.49 \), \( 0.7^4 = 0.49^2 = 0.2401 \), and \( 0.7^5 = 0.2401 \times 0.7 = 0.16807 \). \( A = 60 \times 0.16807 \approx 10.08 \) mg. Check hour by hour: 42, 29.4, 20.58, 14.41, 10.08 mg. The two agree. The model never reaches zero because multiplying a positive amount by 0.7 always leaves a positive amount, however small. The graph approaches the horizontal axis without touching it, which is called a horizontal asymptote. In the body the drug does eventually clear, because the amount becomes smaller than a single molecule and the continuous model stops describing reality, so the model is useful over the first hours and not indefinitely. About 10.08 mg; a positive amount times 0.7 stays positive, so the model approaches zero without reaching it

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What does a constant difference in a table indicate?
   Show the full solution

   A linear function

2. What does a constant ratio indicate?
   Show the full solution

   An exponential function

3. Classify: 3, 7, 11, 15, 19.
   Show the full solution

   Differences are all 4. Linear

4. Classify: 4, 12, 36, 108.
   Show the full solution

   Ratios are all 3. Exponential

5. Is "saves $40 each month" linear or exponential?
   Show the full solution

   A fixed amount is added each step. Linear

6. Classify \( x \): 0, 1, 2, 3 with \( y \): 200, 170, 144.5, 122.825, and write the rule.
   Show the full solution

   The inputs step by 1, so both tests apply. Differences: \( -30 \), \( -25.5 \), \( -21.675 \). Not constant. Ratios: \( 170 \div 200 = 0.85 \), \( 144.5 \div 170 = 0.85 \), \( 122.825 \div 144.5 = 0.85 \). Constant. So it is exponential with \( a = 200 \) and \( b = 0.85 \), a 15 percent decrease each step: \( y = 200(0.85)^x \). Check at \( x = 2 \): \( 200 \times 0.7225 = 144.5 \). Correct. Exponential, \( y = 200(0.85)^x \)

7. Is "a tank loses 8 percent of its contents each hour" linear or exponential? Explain.
   Show the full solution

   Exponential. A percentage is a share of the current amount, and the current amount keeps changing, so the quantity lost each hour is not the same number twice. From 1,000 liters the first hour loses 80 liters, leaving 920; the second hour loses 8 percent of 920, which is 73.6 liters. The multiplier 0.92 is constant while the subtracted amount is not, which is exactly the signature of an exponential. Had the tank lost 80 liters per hour regardless of its contents, that would be linear. Exponential; a constant percentage means a constant ratio, not a constant difference

8. Classify \( x \): 0, 1, 2, 3 with \( y \): 1, 4, 9, 16.
   Show the full solution

   Differences: 3, 5, 7. Not constant, so not linear. Ratios: 4, 2.25, \( 16 \div 9 \approx 1.78 \). Not constant, so not exponential. It is neither. The differences themselves increase by a constant 2, which signals a quadratic pattern, and indeed \( y = (x+1)^2 \). Checking: at \( x = 3 \), \( 4^2 = 16 \). Correct. Testing both possibilities and reporting neither is the right answer here. Neither; it is quadratic

9. Company A pays $2,000 per month rising $100 per month. Company B pays $2,000 rising 3 percent per month. Compare after 1 year and after 5 years.
   Show the full solution

   Company A is linear: \( A = 2000 + 100m \). Company B is exponential: \( B = 2000(1.03)^m \). After 12 months: \( A = 2000 + 1200 = 3200 \), and \( B = 2000(1.03)^{12} \approx 2000 \times 1.425761 \approx 2851.52 \). Company A pays more. After 60 months: \( A = 2000 + 6000 = 8000 \), and \( B = 2000(1.03)^{60} \approx 2000 \times 5.891603 \approx 11783.21 \). Company B now pays considerably more. The reversal is the general result of this lesson: the linear raise of $100 never changes, while B's raise is 3 percent of a salary that keeps climbing, so B's monthly increase eventually exceeds $100 and keeps growing from there. A pays more at one year ($3,200 against about $2,852); B pays far more at five years (about $11,783 against $8,000)

10. Explain why a graph alone is unreliable for distinguishing the two, and what to do instead.
    Show the full solution

    Over a short window an exponential curve is very close to straight, because a smooth curve looks like its tangent line when you zoom in far enough. A plot of \( y = 100(1.02)^x \) for \( x \) from 0 to 5 rises from 100 to about 110 along a path no eye can distinguish from a line. Scaling makes it worse: a badly chosen vertical scale can flatten a steep exponential or exaggerate a gentle line. The reliable approach is arithmetic on the values. Take evenly spaced inputs, compute successive differences and successive ratios, and see which is constant. That test is exact and does not depend on how the picture was drawn. If neither is constant, the honest conclusion is that the data is neither linear nor exponential. Short stretches of an exponential look straight, so compute differences and ratios on evenly spaced inputs instead

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. In \( y = a \cdot b^x \), which parameter is the value at \( x = 0 \)?
   Show the full solution

   \( a \), since \( b^0 = 1 \)

2. A model passes through \( (0, 50) \) and has \( b = 2 \). Write it.
   Show the full solution

   \( y = 50 \cdot 2^x \)

3. What percentage change does \( b = 1.25 \) represent?
   Show the full solution

   A 25 percent increase per step

4. What percentage change does \( b = 0.8 \) represent?
   Show the full solution

   A 20 percent decrease per step

5. Find \( b \) if \( a = 10 \) and the value at \( x = 2 \) is 90.
   Show the full solution

   \( 10b^2 = 90 \), so \( b^2 = 9 \) and \( b = 3 \). \( b = 3 \)

6. A value is 500 at year 0 and 800 at year 4. Build the model and state the annual percentage growth.
   Show the full solution

   Since the first point is at \( x = 0 \), \( a = 500 \). Substituting the second: \( 500b^4 = 800 \), so \( b^4 = 1.6 \) and \( b = 1.6^{1/4} \). Computing: \( 1.6^{1/2} \approx 1.264911 \), and taking the square root again gives \( b \approx 1.124683 \). So \( y = 500(1.1247)^t \) approximately, which is about 12.5 percent growth per year. Check at \( t = 4 \): \( 500 \times 1.124683^4 = 500 \times 1.6 = 800 \). Correct. \( y \approx 500(1.125)^t \), about 12.5 percent per year

7. A quantity is 96 at \( x = 1 \) and 6 at \( x = 5 \). Find \( b \).
   Show the full solution

   Neither point is at \( x = 0 \), so divide to eliminate \( a \): \( \dfrac{ab^5}{ab^1} = \dfrac{6}{96} \), giving \( b^4 = 0.0625 \). Taking the fourth root: \( b^2 = 0.25 \), so \( b = 0.5 \). Check by listing from \( x = 1 \): 96, 48, 24, 12, 6. The value at \( x = 5 \) is 6. Correct. The quantity halves each step, a 50 percent decrease. \( b = 0.5 \)

8. For that quantity, find \( a \) and write the full model.
   Show the full solution

   Using \( ab^1 = 96 \) with \( b = 0.5 \): \( 0.5a = 96 \), so \( a = 192 \). The model is \( y = 192(0.5)^x \). Check at \( x = 1 \): \( 192 \times 0.5 = 96 \). Correct. At \( x = 5 \): \( 192 \times 0.03125 = 6 \). Correct. \( y = 192(0.5)^x \)

9. A town of 5,000 grows 4 percent per year. A student predicts the population in 200 years. Comment on the prediction.
   Show the full solution

   The model is \( P = 5000(1.04)^t \). At \( t = 200 \), \( 1.04^{200} \) is roughly 2,550, so the model predicts a population near 12.7 million in a town that currently holds 5,000 people. The arithmetic is correct and the prediction is not credible. An exponential model assumes the percentage growth rate holds indefinitely, and no town sustains 4 percent annual growth for two centuries: land, water, housing and employment all impose limits, and growth rates fall long before those limits are reached. The model is reasonable over a span where the assumption is plausible, perhaps ten or twenty years, and unreasonable far beyond it. Reporting the number without that caveat would be the actual error. The arithmetic is right but the extrapolation is not credible; the constant growth rate cannot hold for 200 years

10. Data: \( x \): 0, 2, 4, 6 with \( y \): 12, 27, 60.75, 136.6875. Fit a model, verify at every point, and interpret.
    Show the full solution

    The inputs step by 2 rather than 1, so care is needed. Ratios of successive values: \( 27 \div 12 = 2.25 \), \( 60.75 \div 27 = 2.25 \), \( 136.6875 \div 60.75 = 2.25 \). Constant, so the data is exponential. That 2.25 is the factor per two units of \( x \), not per one. Since \( b^2 = 2.25 \), the per-unit factor is \( b = 1.5 \). The value at \( x = 0 \) is 12, so \( a = 12 \) and the model is \( y = 12(1.5)^x \). Verify at every point: \( x = 0 \) gives 12; \( x = 2 \) gives \( 12 \times 2.25 = 27 \); \( x = 4 \) gives \( 12 \times 5.0625 = 60.75 \); \( x = 6 \) gives \( 12 \times 11.390625 = 136.6875 \). All four match exactly. Interpretation: the quantity starts at 12 and increases by 50 percent for each unit increase in \( x \), which is the same as multiplying by 2.25 every two units. The model assumes that rate continues, and it should be trusted only over the range the data covers and a modest distance beyond. \( y = 12(1.5)^x \); 50 percent growth per unit, and the step size of 2 had to be accounted for

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the degree of \( 4x^3 - 2x + 9 \)?
   Show the full solution

   3

2. What is the leading coefficient of \( -6x^2 + x - 4 \)?
   Show the full solution

   The sign belongs to the coefficient. \( -6 \)

3. Add \( (2x + 5) + (3x - 8) \).
   Show the full solution

   \( 5x - 3 \)

4. Subtract \( (7x - 2) - (3x + 4) \).
   Show the full solution

   \( 7x - 2 - 3x - 4 \). \( 4x - 6 \)

5. How many terms does a trinomial have?
   Show the full solution

   Three

6. Write \( 3 - 5x^2 + 8x^4 - x \) in standard form and name its degree and leading coefficient.
   Show the full solution

   Ordering by decreasing exponent: \( 8x^4 - 5x^2 - x + 3 \). The highest exponent is 4, so the degree is 4, and the coefficient on that term is 8. There is no \( x^3 \) term, which simply means its coefficient is zero. \( 8x^4 - 5x^2 - x + 3 \); degree 4, leading coefficient 8

7. Simplify \( (4x^2 - 3x + 1) - (2x^2 - 7x + 5) \).
   Show the full solution

   Distribute the minus to all three terms: \( 4x^2 - 3x + 1 - 2x^2 + 7x - 5 \). Combine: \( 2x^2 + 4x - 4 \). Check at \( x = 2 \): the first polynomial is \( 16 - 6 + 1 = 11 \), the second is \( 8 - 14 + 5 = -1 \), and \( 11 - (-1) = 12 \). The answer gives \( 8 + 8 - 4 = 12 \). They agree. \( 2x^2 + 4x - 4 \)

8. Simplify \( (x^3 + 2x - 6) + (4x^2 - 2x + 6) - (x^3 - x^2) \).
   Show the full solution

   Handle the signs first: \( x^3 + 2x - 6 + 4x^2 - 2x + 6 - x^3 + x^2 \). Combine by degree: \( x^3 \): \( x^3 - x^3 = 0 \). \( x^2 \): \( 4x^2 + x^2 = 5x^2 \). \( x \): \( 2x - 2x = 0 \). Constants: \( -6 + 6 = 0 \). The result is \( 5x^2 \). Check at \( x = 2 \): \( (8 + 4 - 6) + (16 - 4 + 6) - (8 - 4) = 6 + 18 - 4 = 20 \), and \( 5(4) = 20 \). Correct. \( 5x^2 \)

9. Explain why \( 3x^{-2} + 5 \) is not a polynomial.
   Show the full solution

   Every term of a polynomial must have the variable raised to a whole-number power. Here the exponent is \( -2 \), and by the definition from lesson 6.4 that term is \( \dfrac{3}{x^2} \), a variable in a denominator. That is a rational expression, not a polynomial term. The distinction matters because the whole-number requirement is what guarantees polynomials behave well: they are defined for every value of \( x \), they have no breaks in their graphs, and the factoring methods in this unit apply to them. An expression with \( x \) in a denominator is undefined at \( x = 0 \) and needs the separate treatment given in unit 10. The exponent \( -2 \) is not a whole number, so the term has the variable in a denominator

10. A rectangle has length \( 2x + 5 \) and width \( x - 3 \). Write its perimeter, evaluate at \( x = 7 \), and state the restriction on \( x \).
    Show the full solution

    Perimeter is twice the length plus twice the width: \( P = 2(2x + 5) + 2(x - 3) = 4x + 10 + 2x - 6 = 6x + 4 \). At \( x = 7 \): \( P = 42 + 4 = 46 \) units. Checking directly, the length is \( 2(7) + 5 = 19 \) and the width is \( 7 - 3 = 4 \), giving a perimeter of \( 2(19) + 2(4) = 38 + 8 = 46 \). They agree. The restriction comes from the width. A side length must be positive, so \( x - 3 \gt 0 \), giving \( x \gt 3 \). The length \( 2x + 5 \) is automatically positive when \( x \gt 3 \), so the width provides the binding constraint. Without this restriction the algebra would happily accept \( x = 1 \) and report a perimeter of 10 for a rectangle with a width of \( -2 \). \( P = 6x + 4 \), which is 46 at \( x = 7 \); \( x \gt 3 \) is required

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Expand \( 4x(x + 3) \).
   Show the full solution

   \( 4x^2 + 12x \)

2. Expand \( -2x(3x - 5) \).
   Show the full solution

   \( -2x \cdot 3x = -6x^2 \) and \( -2x \cdot (-5) = 10x \). \( -6x^2 + 10x \)

3. Expand \( (x + 4)(x + 6) \).
   Show the full solution

   \( x^2 + 6x + 4x + 24 \). \( x^2 + 10x + 24 \)

4. Expand \( (x - 3)(x + 7) \).
   Show the full solution

   \( x^2 + 7x - 3x - 21 \). \( x^2 + 4x - 21 \)

5. How many products come from a binomial times a trinomial?
   Show the full solution

   Six

6. Expand \( (3x - 4)(2x + 5) \) and check numerically.
   Show the full solution

   \( 3x(2x + 5) - 4(2x + 5) = 6x^2 + 15x - 8x - 20 = 6x^2 + 7x - 20 \). Check at \( x = 2 \): the original is \( (2)(9) = 18 \), and the answer is \( 24 + 14 - 20 = 18 \). Correct. \( 6x^2 + 7x - 20 \)

7. Expand \( (2x - 1)(x^2 + 3x - 2) \).
   Show the full solution

   Six products expected. \( 2x(x^2 + 3x - 2) = 2x^3 + 6x^2 - 4x \). \( -1(x^2 + 3x - 2) = -x^2 - 3x + 2 \). Combining: \( 2x^3 + 5x^2 - 7x + 2 \). Degree check: \( 1 + 2 = 3 \). Correct. Value check at \( x = 1 \): the original is \( (1)(2) = 2 \), and the answer is \( 2 + 5 - 7 + 2 = 2 \). Correct. \( 2x^3 + 5x^2 - 7x + 2 \)

8. Expand \( (x + 1)(x + 2)(x + 3) \).
   Show the full solution

   Multiply two factors first: \( (x + 1)(x + 2) = x^2 + 3x + 2 \). Then multiply by the third: \( (x^2 + 3x + 2)(x + 3) = x(x^2 + 3x + 2) + 3(x^2 + 3x + 2) \) \( = x^3 + 3x^2 + 2x + 3x^2 + 9x + 6 = x^3 + 6x^2 + 11x + 6 \). Degree check: \( 1 + 1 + 1 = 3 \). Correct. Value check at \( x = 1 \): the original is \( 2 \times 3 \times 4 = 24 \), and the answer is \( 1 + 6 + 11 + 6 = 24 \). Correct. \( x^3 + 6x^2 + 11x + 6 \)

9. A student expands \( (x - 4)(x - 6) \) as \( x^2 - 10x - 24 \). Find and explain the error.
   Show the full solution

   The four products are \( x \cdot x = x^2 \), \( x \cdot (-6) = -6x \), \( -4 \cdot x = -4x \) and \( -4 \cdot (-6) = +24 \). The last one is where they slipped: a negative times a negative is positive, so the constant is \( +24 \), not \( -24 \). The correct expansion is \( x^2 - 10x + 24 \). A numerical check settles it instantly: at \( x = 0 \) the original is \( (-4)(-6) = 24 \), so the constant term must be \( +24 \). Substituting zero is a fast way to verify the constant term of any expansion. \( x^2 - 10x + 24 \); the product of the two negative constants is positive

10. A rectangle has length \( x + 7 \) and width \( x + 2 \). Write its area, then find the area if both dimensions increase by 3.
    Show the full solution

    Area is length times width: \( A = (x + 7)(x + 2) = x^2 + 2x + 7x + 14 = x^2 + 9x + 14 \). Increasing both dimensions by 3 gives a length of \( x + 10 \) and a width of \( x + 5 \): \( A_{\text{new}} = (x + 10)(x + 5) = x^2 + 5x + 10x + 50 = x^2 + 15x + 50 \). The increase is \( (x^2 + 15x + 50) - (x^2 + 9x + 14) = 6x + 36 \), which is not a constant. That is worth noticing: adding 3 to each dimension adds more area to a large rectangle than to a small one, since the added strips are longer. Check at \( x = 1 \): the original is \( 8 \times 3 = 24 \), the new one is \( 11 \times 6 = 66 \), and the difference is 42. The expression \( 6x + 36 \) gives \( 6 + 36 = 42 \). Correct. \( A = x^2 + 9x + 14 \) and \( A_{\text{new}} = x^2 + 15x + 50 \), an increase of \( 6x + 36 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Expand \( (x + 3)^2 \).
   Show the full solution

   \( x^2 + 6x + 9 \)

2. Expand \( (x - 7)^2 \).
   Show the full solution

   \( x^2 - 14x + 49 \)

3. Expand \( (x + 6)(x - 6) \).
   Show the full solution

   \( x^2 - 36 \)

4. Expand \( (5x)^2 \).
   Show the full solution

   \( 25x^2 \)

5. Why does \( (a+b)(a-b) \) have no middle term?
   Show the full solution

   The products \( +ab \) and \( -ab \) cancel

6. Expand \( (4x + 3)^2 \) and check numerically.
   Show the full solution

   With \( a = 4x \) and \( b = 3 \): \( a^2 = 16x^2 \), \( 2ab = 2(4x)(3) = 24x \), and \( b^2 = 9 \). So \( (4x + 3)^2 = 16x^2 + 24x + 9 \). Check at \( x = 1 \): the original is \( 7^2 = 49 \), and the answer is \( 16 + 24 + 9 = 49 \). Correct. \( 16x^2 + 24x + 9 \)

7. Expand \( (3x - 5y)(3x + 5y) \).
   Show the full solution

   The binomials differ only in sign, so the difference of squares applies with \( a = 3x \) and \( b = 5y \): \( (3x)^2 - (5y)^2 = 9x^2 - 25y^2 \). Check at \( x = 1 \), \( y = 1 \): the original is \( (-2)(8) = -16 \), and the answer is \( 9 - 25 = -16 \). Correct. \( 9x^2 - 25y^2 \)

8. Is \( x^2 + 12x + 36 \) a perfect square trinomial? Justify.
   Show the full solution

   Test the three conditions. The first term \( x^2 \) is a perfect square with root \( x \). The last term 36 is a perfect square with root 6. The middle term must be twice the product of those roots: \( 2(x)(6) = 12x \), which is exactly what is there. So yes, and it factors as \( (x + 6)^2 \). Checking by expanding: \( x^2 + 12x + 36 \). Correct. Had the middle term been anything else, such as \( 13x \), the trinomial would not be a perfect square, even though both outer terms are. Yes; it is \( (x + 6)^2 \)

9. Explain why \( (a + b)^2 \neq a^2 + b^2 \) using both algebra and a numerical example.
   Show the full solution

   Algebraically, \( (a + b)^2 \) means \( (a + b)(a + b) \), and distributing gives four products: \( a^2 \), \( ab \), \( ba \) and \( b^2 \). The two cross products are equal, so they combine into \( 2ab \), and the correct expansion is \( a^2 + 2ab + b^2 \). Writing \( a^2 + b^2 \) discards \( 2ab \) entirely. The underlying mistake is treating the exponent as distributing over addition. Exponents distribute over multiplication, so \( (ab)^2 = a^2 b^2 \) is true, but nothing similar holds for sums. Numerically, take \( a = 3 \) and \( b = 4 \): \( (3 + 4)^2 = 49 \), while \( 3^2 + 4^2 = 25 \). The difference of 24 is precisely \( 2ab = 2(3)(4) \). The expansion is \( a^2 + 2ab + b^2 \); the \( 2ab \) is dropped, as \( (3+4)^2 = 49 \) against \( 9 + 16 = 25 \) shows

10. Use the difference of squares to compute \( 103 \times 97 \) mentally, and explain the method.
    Show the full solution

    Both numbers sit 3 away from 100, one above and one below, so they can be written as \( 100 + 3 \) and \( 100 - 3 \). That is exactly the difference of squares pattern with \( a = 100 \) and \( b = 3 \): \( (100 + 3)(100 - 3) = 100^2 - 3^2 = 10000 - 9 = 9991 \). The method works whenever two numbers are equally spaced around a round number, since their product is then the square of that round number minus the square of the gap. For instance \( 48 \times 52 = 50^2 - 2^2 = 2500 - 4 = 2496 \). Verifying by long multiplication: \( 103 \times 97 = 103 \times 100 - 103 \times 3 = 10300 - 309 = 9991 \). Correct. 9,991, since \( 103 \times 97 = 100^2 - 3^2 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Factor \( 6x + 9 \).
   Show the full solution

   \( 3(2x + 3) \)

2. Factor \( 5x^2 - 15x \).
   Show the full solution

   \( 5x(x - 3) \)

3. What is the GCF of \( x^5 \) and \( x^2 \)?
   Show the full solution

   The smaller exponent. \( x^2 \)

4. Factor \( 8a^3 + 12a^2 \).
   Show the full solution

   GCF is \( 4a^2 \). \( 4a^2(2a + 3) \)

5. How do you check a factorization?
   Show the full solution

   Multiply the factors back out and compare with the original

6. Factor \( 24x^4 y - 36x^3 y^2 + 12x^2 y \).
   Show the full solution

   Coefficients 24, 36 and 12 share a GCF of 12. For \( x \): exponents 4, 3, 2, so \( x^2 \). For \( y \): exponents 1, 2, 1, so \( y \). The GCF is \( 12x^2 y \). Dividing: \( 2x^2 \), \( -3xy \), and \( 1 \). The last term becomes 1, not nothing. So the factorization is \( 12x^2 y(2x^2 - 3xy + 1) \). Check: \( 12x^2y \cdot 2x^2 = 24x^4y \); \( 12x^2y \cdot (-3xy) = -36x^3y^2 \); \( 12x^2y \cdot 1 = 12x^2y \). All recovered. \( 12x^2 y(2x^2 - 3xy + 1) \)

7. Factor \( -4x^3 + 10x^2 - 2x \), pulling out a negative.
   Show the full solution

   The numeric GCF is 2 and the variable GCF is \( x \), so with the negative the factor is \( -2x \). Dividing: \( \dfrac{-4x^3}{-2x} = 2x^2 \); \( \dfrac{10x^2}{-2x} = -5x \); \( \dfrac{-2x}{-2x} = 1 \). So \( -4x^3 + 10x^2 - 2x = -2x(2x^2 - 5x + 1) \). Check: \( -2x \cdot 2x^2 = -4x^3 \); \( -2x \cdot (-5x) = 10x^2 \); \( -2x \cdot 1 = -2x \). Correct. \( -2x(2x^2 - 5x + 1) \)

8. Factor \( 3x(x - 4) + 7(x - 4) \).
   Show the full solution

   The common factor here is not a monomial but the whole binomial \( (x - 4) \), which appears in both terms. Treating it as a single object and pulling it out: \( 3x(x - 4) + 7(x - 4) = (x - 4)(3x + 7) \). Check by distributing: \( (x-4)(3x+7) = 3x^2 + 7x - 12x - 28 = 3x^2 - 5x - 28 \), and the original expands to \( 3x^2 - 12x + 7x - 28 = 3x^2 - 5x - 28 \). They agree. This is the step that makes factoring by grouping work in lesson 7.6. \( (x - 4)(3x + 7) \)

9. A student factors \( 7x^2 + 7x + 7 \) as \( 7(x^2 + x) \). Find the error.
   Show the full solution

   They lost the third term. Dividing each term by 7 gives \( x^2 \), \( x \) and 1, so the correct factorization is \( 7(x^2 + x + 1) \). The final term divided to 1 rather than disappearing, which is the same slip as writing nothing where a quotient of 1 belongs. Multiplying back exposes it immediately: \( 7(x^2 + x) = 7x^2 + 7x \), which is missing the \( +7 \). Every factorization should be checked this way, and a bracket with fewer terms than the original is always a warning sign. \( 7(x^2 + x + 1) \); the constant term divided to 1, not to nothing

10. Factor \( 2x^3 y^2 + 8x^2 y^3 - 6xy^4 \) completely, and explain why factoring out the GCF first is worth doing.
    Show the full solution

    Coefficients 2, 8 and 6 share a GCF of 2. For \( x \): exponents 3, 2, 1, so \( x \). For \( y \): exponents 2, 3, 4, so \( y^2 \). The GCF is \( 2xy^2 \). Dividing: \( \dfrac{2x^3y^2}{2xy^2} = x^2 \); \( \dfrac{8x^2y^3}{2xy^2} = 4xy \); \( \dfrac{-6xy^4}{2xy^2} = -3y^2 \). So the factorization is \( 2xy^2(x^2 + 4xy - 3y^2) \). Check: \( 2xy^2 \cdot x^2 = 2x^3y^2 \); \( 2xy^2 \cdot 4xy = 8x^2y^3 \); \( 2xy^2 \cdot (-3y^2) = -6xy^4 \). All three recovered. Pulling out the GCF first is worth doing for two reasons. Practically, whatever method comes next works on smaller coefficients and lower exponents, so there is less to go wrong. Formally, a factorization is only complete when no factor can be broken down further, so leaving a common factor inside means the answer is not finished even if every other step was right. \( 2xy^2(x^2 + 4xy - 3y^2) \); the GCF makes later steps smaller and a factorization is incomplete without it

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Factor \( x^2 + 5x + 6 \).
   Show the full solution

   2 and 3 multiply to 6 and add to 5. \( (x + 2)(x + 3) \)

2. Factor \( x^2 + 9x + 20 \).
   Show the full solution

   \( (x + 4)(x + 5) \)

3. Factor \( x^2 - 7x + 10 \).
   Show the full solution

   Both negative: \( -2 \) and \( -5 \). \( (x - 2)(x - 5) \)

4. Factor \( x^2 + 2x - 8 \).
   Show the full solution

   Opposite signs: 4 and \( -2 \). \( (x + 4)(x - 2) \)

5. If \( c \) is negative, what do you know about the two numbers?
   Show the full solution

   They have opposite signs

6. Factor \( x^2 - 3x - 40 \).
   Show the full solution

   The constant is negative, so the numbers have opposite signs, and the middle term is negative, so the larger one is negative. Pairs multiplying to \( -40 \): 1 and \( -40 \) sum to \( -39 \); 2 and \( -20 \) sum to \( -18 \); 4 and \( -10 \) sum to \( -6 \); 5 and \( -8 \) sum to \( -3 \). The last matches. So \( x^2 - 3x - 40 = (x + 5)(x - 8) \). Check: \( x^2 - 8x + 5x - 40 = x^2 - 3x - 40 \). Correct. \( (x + 5)(x - 8) \)

7. Factor \( 3x^2 + 21x + 36 \) completely.
   Show the full solution

   Factor out the GCF 3 first: \( 3(x^2 + 7x + 12) \). Now the trinomial has a leading coefficient of 1, and 3 and 4 multiply to 12 and add to 7: \( 3(x + 3)(x + 4) \). Check: \( (x+3)(x+4) = x^2 + 7x + 12 \), and multiplying by 3 gives \( 3x^2 + 21x + 36 \). Correct. Skipping the GCF step would have required factoring a trinomial with leading coefficient 3 by the harder method of lesson 7.6, for no benefit. \( 3(x + 3)(x + 4) \)

8. Factor \( x^2 - 16x + 64 \) and say what is special about it.
   Show the full solution

   Both numbers are negative with product 64 and sum \( -16 \): \( -8 \) and \( -8 \). So \( x^2 - 16x + 64 = (x - 8)(x - 8) = (x - 8)^2 \). What is special is that the two factors are identical, making this a perfect square trinomial from lesson 7.3. It fits that pattern: the first term is \( x^2 \), the last is \( 8^2 \), and the middle is \( 2(x)(8) = 16x \) with a minus sign. Check: \( (x-8)^2 = x^2 - 16x + 64 \). Correct. \( (x - 8)^2 \), a perfect square trinomial

9. Determine whether \( x^2 + 4x + 7 \) factors over the integers, showing the search.
   Show the full solution

   The constant 7 is positive and the middle term is positive, so both numbers must be positive with product 7 and sum 4. Since 7 is prime, the only positive factor pair is 1 and 7, which sums to 8 rather than 4. There are no other integer possibilities to test, so the search is complete and exhaustive. The trinomial is prime over the integers. This does not mean it has no roots at all; it means no factorization with integer coefficients exists. The methods of unit 9 will find its roots by other means, and they turn out not to be rational. It is prime over the integers; the only factor pair 1 and 7 sums to 8

10. Find every integer \( b \) making \( x^2 + bx + 18 \) factorable over the integers.
    Show the full solution

    Factoring requires two integers with product 18, and \( b \) is then their sum. So the question is which sums the factor pairs of 18 can produce. Positive pairs: 1 and 18 sum to 19; 2 and 9 sum to 11; 3 and 6 sum to 9. Negative pairs give the same products with opposite sums: \( -1 \) and \( -18 \) sum to \( -19 \); \( -2 \) and \( -9 \) sum to \( -11 \); \( -3 \) and \( -6 \) sum to \( -9 \). Mixed-sign pairs are impossible here, because the product 18 is positive. So \( b \) can be \( \pm 19 \), \( \pm 11 \) or \( \pm 9 \), and no other value works. Checking one: with \( b = 11 \), \( x^2 + 11x + 18 = (x + 2)(x + 9) \), which expands correctly. \( b = \pm 9, \pm 11, \pm 19 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( 2x^2 + 7x + 3 \), what is \( ac \)?
   Show the full solution

   \( 2 \times 3 \). 6

2. Factor \( 2x^2 + 7x + 3 \).
   Show the full solution

   1 and 6 multiply to 6 and add to 7. Splitting: \( 2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) \). \( (2x + 1)(x + 3) \)

3. Factor \( 3x^2 + 8x + 4 \).
   Show the full solution

   \( ac = 12 \); 2 and 6 add to 8. \( 3x^2 + 2x + 6x + 4 = x(3x+2) + 2(3x+2) \). \( (3x + 2)(x + 2) \)

4. After splitting and grouping, what must be true of the two brackets?
   Show the full solution

   They must be identical

5. Factor \( 5x^2 - 13x + 6 \).
   Show the full solution

   \( ac = 30 \); both negative, \( -3 \) and \( -10 \) add to \( -13 \). \( 5x^2 - 3x - 10x + 6 = x(5x-3) - 2(5x-3) \). \( (5x - 3)(x - 2) \)

6. Factor \( 6x^2 - 7x - 20 \).
   Show the full solution

   \( ac = 6 \times (-20) = -120 \), and the numbers must add to \( -7 \). Opposite signs with the larger one negative. Testing pairs of 120: 8 and 15 differ by 7, so the numbers are \( -15 \) and \( +8 \). Check: \( -15 \times 8 = -120 \) and \( -15 + 8 = -7 \). Correct. Splitting: \( 6x^2 - 15x + 8x - 20 \). Grouping: \( 3x(2x - 5) + 4(2x - 5) \). Factoring: \( (2x - 5)(3x + 4) \). Check: \( 6x^2 + 8x - 15x - 20 = 6x^2 - 7x - 20 \). Correct. \( (2x - 5)(3x + 4) \)

7. Factor \( 12x^2 + 26x + 10 \) completely.
   Show the full solution

   The GCF is 2, so factor it out first: \( 2(6x^2 + 13x + 5) \). This makes the remaining numbers much smaller. For the trinomial, \( ac = 30 \) and the numbers must add to 13, both positive: 3 and 10. Splitting: \( 6x^2 + 3x + 10x + 5 \). Grouping: \( 3x(2x + 1) + 5(2x + 1) = (2x + 1)(3x + 5) \). So the complete factorization is \( 2(2x + 1)(3x + 5) \). Check: \( (2x+1)(3x+5) = 6x^2 + 10x + 3x + 5 = 6x^2 + 13x + 5 \), and doubling gives \( 12x^2 + 26x + 10 \). Correct. \( 2(2x + 1)(3x + 5) \)

8. Factor \( 4x^2 - 12x + 9 \).
   Show the full solution

   \( ac = 36 \) and the numbers must add to \( -12 \), both negative: \( -6 \) and \( -6 \). Splitting: \( 4x^2 - 6x - 6x + 9 \). Grouping: \( 2x(2x - 3) - 3(2x - 3) = (2x - 3)(2x - 3) = (2x - 3)^2 \). The identical factors reveal a perfect square trinomial, which matches the lesson 7.3 pattern: \( (2x)^2 = 4x^2 \), \( 3^2 = 9 \), and \( 2(2x)(3) = 12x \) with a minus. Check: \( (2x-3)^2 = 4x^2 - 12x + 9 \). Correct. \( (2x - 3)^2 \)

9. A student splits \( 6x^2 + 11x - 35 \) correctly but groups as \( 3x(2x+7) + 5(-2x-7) \). Explain what went wrong and fix it.
   Show the full solution

   The split into \( 6x^2 + 21x - 10x - 35 \) was right. The problem is in the second pair: they factored out \( +5 \) from \( -10x - 35 \), which forces the bracket to become \( -2x - 7 \), not matching the first bracket \( 2x + 7 \). With mismatched brackets there is nothing common to pull out and the method stalls. The fix is to factor out \( -5 \) instead: \( -10x - 35 = -5(2x + 7) \). Now both brackets read \( (2x + 7) \) and the common factor comes out: \( (2x + 7)(3x - 5) \). The general rule is that when the second pair begins with a negative term, the negative belongs in the factor. Their version is not wrong in value, since \( 5(-2x - 7) = -10x - 35 \) is correct arithmetic, but it is written in a form the method cannot continue from. Factor out \( -5 \), not \( +5 \), giving \( (2x + 7)(3x - 5) \)

10. Determine whether \( 4x^2 + 5x + 3 \) factors over the integers, showing the complete search.
    Show the full solution

    There is no GCF, since 4, 5 and 3 share only 1. \( ac = 4 \times 3 = 12 \), and the numbers must add to 5. Both the product and the sum are positive, so both numbers are positive. The positive factor pairs of 12 are: 1 and 12, summing to 13; 2 and 6, summing to 8; 3 and 4, summing to 7. No other pairs exist. None of these sums is 5, and the search is exhaustive, so no integer split of the middle term exists and the trinomial is prime over the integers. As in lesson 7.5, prime is a complete answer. It does not mean the expression has no roots, only that it has no factorization with integer coefficients, and the methods of unit 9 will show that its roots are not real numbers at all. Prime over the integers; no factor pair of 12 sums to 5

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Factor \( x^2 - 49 \).
   Show the full solution

   \( (x + 7)(x - 7) \)

2. Factor \( 4x^2 - 81 \).
   Show the full solution

   \( (2x + 9)(2x - 9) \)

3. Does \( x^2 + 9 \) factor over the integers?
   Show the full solution

   No; a sum of squares is prime

4. What is the first step of every factoring problem?
   Show the full solution

   Factor out the greatest common factor

5. Factor \( 5x^2 - 20 \).
   Show the full solution

   GCF 5 first, then the pattern: \( 5(x^2 - 4) \). \( 5(x + 2)(x - 2) \)

6. Factor \( 8x^3 - 50x \) completely.
   Show the full solution

   GCF first: the terms share \( 2x \), giving \( 2x(4x^2 - 25) \). The bracket has two terms, subtraction, and both are perfect squares: \( 4x^2 = (2x)^2 \) and \( 25 = 5^2 \). So \( 8x^3 - 50x = 2x(2x + 5)(2x - 5) \). Check at \( x = 1 \): the original is \( 8 - 50 = -42 \), and the factorization gives \( 2(7)(-3) = -42 \). Correct. \( 2x(2x + 5)(2x - 5) \)

7. Factor \( x^4 - 81 \) completely.
   Show the full solution

   Both terms are perfect squares: \( x^4 = (x^2)^2 \) and \( 81 = 9^2 \). First round: \( x^4 - 81 = (x^2 + 9)(x^2 - 9) \). Examine each factor. \( x^2 + 9 \) is a sum of squares and is prime. \( x^2 - 9 \) is a difference of squares and factors as \( (x + 3)(x - 3) \). So the complete factorization is \( (x^2 + 9)(x + 3)(x - 3) \). Check at \( x = 2 \): the original is \( 16 - 81 = -65 \), and the factorization gives \( (13)(5)(-1) = -65 \). Correct. \( (x^2 + 9)(x + 3)(x - 3) \)

8. Factor \( 2x^3 + 6x^2 - 20x \) completely.
   Show the full solution

   GCF first: all three terms share \( 2x \), giving \( 2x(x^2 + 3x - 10) \). The bracket is a trinomial with leading coefficient 1, so find two numbers multiplying to \( -10 \) and adding to 3: 5 and \( -2 \). So \( 2x^3 + 6x^2 - 20x = 2x(x + 5)(x - 2) \). Check at \( x = 1 \): the original is \( 2 + 6 - 20 = -12 \), and the factorization gives \( 2(6)(-1) = -12 \). Correct. This problem needed two different methods in sequence, which is why the checklist says to re-examine each factor after every step. \( 2x(x + 5)(x - 2) \)

9. A student factors \( 3x^4 - 48 \) as \( 3(x^2 + 4)(x^2 - 4) \) and stops. Complete it and explain the principle.
   Show the full solution

   Every step so far is correct, but the work is unfinished. Factoring completely means continuing until no factor can be broken down further, and \( x^2 - 4 \) is still a difference of squares, factoring as \( (x + 2)(x - 2) \). The complete answer is \( 3(x^2 + 4)(x + 2)(x - 2) \). The factor \( x^2 + 4 \) is a sum of squares and genuinely cannot be factored over the integers, so the work stops there. The principle is that after every round of factoring, each new factor must be examined afresh. A single application of a pattern rarely finishes a problem, and the habit of looking again is what separates a complete factorization from a partial one. \( 3(x^2 + 4)(x + 2)(x - 2) \); re-examine every new factor until none factors further

10. Factor \( x^3 + 3x^2 - 4x - 12 \) completely by grouping.
    Show the full solution

    There is no common factor across all four terms, so move to the four-term entry on the checklist, which is grouping. Group in pairs: \( (x^3 + 3x^2) + (-4x - 12) \). Factor each pair: \( x^2(x + 3) - 4(x + 3) \). The negative came out of the second pair so that the brackets match. Factor out the common bracket: \( (x + 3)(x^2 - 4) \). Now examine the new factors. \( x^2 - 4 \) is a difference of squares, so it factors further: \( (x + 2)(x - 2) \). The complete factorization is \( (x + 3)(x + 2)(x - 2) \). Check at \( x = 1 \): the original is \( 1 + 3 - 4 - 12 = -12 \), and the factorization gives \( (4)(3)(-1) = -12 \). Correct. \( (x + 3)(x + 2)(x - 2) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the vertex of \( y = (x - 4)^2 \)?
   Show the full solution

   \( (4, 0) \)

2. What is the vertex of \( y = x^2 + 7 \)?
   Show the full solution

   \( (0, 7) \)

3. Does \( y = -3x^2 \) open up or down?
   Show the full solution

   Down, since \( a \) is negative

4. Is \( y = \frac{1}{4}x^2 \) wider or narrower than the parent?
   Show the full solution

   Wider, since the size of \( a \) is less than 1

5. What is the vertex of \( y = 2(x + 5)^2 - 1 \)?
   Show the full solution

   The bracket is zero at \( x = -5 \). \( (-5, -1) \)

6. Describe every transformation in \( y = \frac{1}{2}(x + 3)^2 - 4 \).
   Show the full solution

   Matching the form: \( a = \frac{1}{2} \), \( h = -3 \) and \( k = -4 \). The size of \( a \) is less than 1, so the parabola is wider than the parent, and \( a \) is positive, so it opens upward. The bracket \( (x + 3) \) is zero at \( x = -3 \), so the graph shifts 3 units left. The \( -4 \) shifts it 4 units down. The vertex is \( (-3, -4) \), and it is the lowest point. Wider by a factor of \( \frac{1}{2} \), opens upward, shifted 3 left and 4 down, vertex \( (-3, -4) \)

7. Write the equation of a parabola with vertex \( (2, -7) \) opening upward, congruent to the parent.
   Show the full solution

   Congruent to the parent means the same width, so \( a = 1 \). Opening upward confirms \( a \) is positive. With \( h = 2 \) and \( k = -7 \): \( y = (x - 2)^2 - 7 \). Check that the vertex is right: at \( x = 2 \), \( y = 0 - 7 = -7 \). Correct. \( y = (x - 2)^2 - 7 \)

8. Find the point 3 units right of the vertex of \( y = -(x - 1)^2 + 6 \).
   Show the full solution

   The vertex is \( (1, 6) \). Moving 3 units right means \( x = 4 \). On the parent, moving 3 from the vertex changes the height by \( 3^2 = 9 \). Here \( a = -1 \), so the change is \( -9 \), and the height is \( 6 - 9 = -3 \). Verify by substituting: \( y = -(4 - 1)^2 + 6 = -9 + 6 = -3 \). Correct. \( (4, -3) \)

9. Explain why \( (x - h)^2 \) shifts right when \( h \) is positive, which looks backwards.
   Show the full solution

   The vertex occurs where the squared quantity is zero, since a square is never negative and zero is its smallest possible value. For \( (x - h)^2 \) that happens when \( x = h \), so the turning point sits at \( x = h \), to the right of the origin when \( h \) is positive. Put another way, the graph does at \( x = h \) whatever the parent did at \( x = 0 \), because the bracket feeds the parent function the value \( x - h \) rather than \( x \). To get the same input the parent had at zero, the new \( x \) must be larger by \( h \), and larger means further right. A concrete check: \( y = (x - 3)^2 \) at \( x = 3 \) gives 0, the same value the parent gives at \( x = 0 \). The whole picture has slid 3 to the right. The vertex sits where the bracket is zero, which is \( x = h \)

10. Two parabolas have vertex \( (0,0) \), one with \( a = 3 \) and one with \( a = \frac{1}{3} \). Compare their heights at \( x = 2 \) and \( x = 6 \), and describe the effect of \( a \).
    Show the full solution

    At \( x = 2 \): the parent height is \( 2^2 = 4 \). The first gives \( 3 \times 4 = 12 \); the second gives \( \frac{1}{3} \times 4 \approx 1.33 \). At \( x = 6 \): the parent height is 36. The first gives 108; the second gives 12. The number \( a \) multiplies every height of the parent parabola. With \( a = 3 \) the graph climbs three times as fast, so it is squeezed toward the axis of symmetry and looks narrow. With \( a = \frac{1}{3} \) the graph climbs a third as fast, so it spreads out and looks wide. A useful way to see it: the first parabola reaches a height of 12 by \( x = 2 \), while the second needs \( x = 6 \) to get there. Both are genuine parabolas of the same family; only the vertical scale differs. 12 and 1.33 at \( x = 2 \); 108 and 12 at \( x = 6 \). The value \( a \) scales every height, making the graph narrow or wide

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Give the vertex of \( y = 3(x - 5)^2 + 2 \).
   Show the full solution

   \( (5, 2) \)

2. Give the axis of symmetry of \( y = -(x + 4)^2 \).
   Show the full solution

   \( x = -4 \)

3. Find the \( y \) intercept of \( y = (x - 3)^2 + 1 \).
   Show the full solution

   At \( x = 0 \): \( 9 + 1 = 10 \). \( (0, 10) \)

4. Does \( y = 4(x - 1)^2 + 3 \) have \( x \) intercepts?
   Show the full solution

   The minimum value is 3, above the axis, and the parabola opens upward. No

5. What is the minimum value of \( y = 2(x + 6)^2 - 11 \)?
   Show the full solution

   \( -11 \), at \( x = -6 \)

6. Find both \( x \) intercepts of \( y = (x - 2)^2 - 9 \).
   Show the full solution

   Set \( y = 0 \): \( (x - 2)^2 = 9 \). Taking both roots: \( x - 2 = 3 \) gives \( x = 5 \); \( x - 2 = -3 \) gives \( x = -1 \). Check the symmetry: the midpoint of 5 and \( -1 \) is 2, which is the axis. Correct. \( (5, 0) \) and \( (-1, 0) \)

7. Write the equation with vertex \( (-1, 4) \) passing through \( (1, 0) \).
   Show the full solution

   The form is \( y = a(x + 1)^2 + 4 \). Substituting \( (1, 0) \): \( 0 = a(2)^2 + 4 \), so \( 4a = -4 \) and \( a = -1 \). The equation is \( y = -(x + 1)^2 + 4 \). Check: at \( x = -1 \) the value is 4, and at \( x = 1 \) it is \( -4 + 4 = 0 \). Both correct. The negative \( a \) means the parabola opens downward, which makes sense since the given point is below the vertex. \( y = -(x + 1)^2 + 4 \)

8. Find the \( x \) intercepts of \( y = 3(x - 1)^2 - 12 \).
   Show the full solution

   Set \( y = 0 \): \( 3(x - 1)^2 = 12 \), so \( (x - 1)^2 = 4 \). Both roots: \( x - 1 = 2 \) gives \( x = 3 \); \( x - 1 = -2 \) gives \( x = -1 \). Check: at \( x = 3 \), \( 3(4) - 12 = 0 \). Correct. The midpoint of 3 and \( -1 \) is 1, matching the axis. \( (3, 0) \) and \( (-1, 0) \)

9. Explain how to tell from vertex form alone how many \( x \) intercepts a parabola has.
   Show the full solution

   Compare the sign of \( a \) with the sign of \( k \), since \( k \) is the height of the vertex. If \( a \) and \( k \) have opposite signs, the vertex is on the opposite side of the axis from the direction the parabola opens, so the curve must cross twice. For example \( y = 2(x-1)^2 - 5 \) opens upward from a vertex below the axis. If they have the same sign, the parabola opens away from the axis and never reaches it, giving no \( x \) intercepts. For example \( y = 2(x-1)^2 + 5 \) has a lowest point at height 5 and climbs from there. If \( k = 0 \), the vertex sits on the axis and the parabola touches it at exactly one point, giving a single intercept. Opposite signs give two intercepts, the same sign gives none, and \( k = 0 \) gives exactly one

10. A parabola has \( x \) intercepts at \( -4 \) and 2 and passes through \( (0, -8) \). Find its vertex form.
    Show the full solution

    The axis of symmetry lies halfway between the intercepts, since a parabola is symmetric: \( h = \dfrac{-4 + 2}{2} = -1 \). So the form is \( y = a(x + 1)^2 + k \), with two unknowns and two usable facts. Substituting the intercept \( (2, 0) \): \( 0 = a(3)^2 + k \), so \( 9a + k = 0 \). Substituting the point \( (0, -8) \): \( -8 = a(1)^2 + k \), so \( a + k = -8 \). Subtracting the second equation from the first: \( 8a = 8 \), so \( a = 1 \), and then \( k = -9 \). The equation is \( y = (x + 1)^2 - 9 \). Check all three given facts. At \( x = -4 \): \( 9 - 9 = 0 \). At \( x = 2 \): \( 9 - 9 = 0 \). At \( x = 0 \): \( 1 - 9 = -8 \). All correct. \( y = (x + 1)^2 - 9 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the axis of symmetry of \( y = x^2 + 4x + 1 \).
   Show the full solution

   \( -\frac{4}{2} = -2 \). \( x = -2 \)

2. Find the \( y \) intercept of \( y = 2x^2 - 3x + 7 \).
   Show the full solution

   \( (0, 7) \)

3. Does \( y = -x^2 + 2x + 3 \) have a maximum or a minimum?
   Show the full solution

   A maximum, since \( a \) is negative

4. Find the axis of \( y = 2x^2 - 8x + 1 \).
   Show the full solution

   \( -\frac{-8}{4} = 2 \). \( x = 2 \)

5. After finding the axis, how do you get the vertex?
   Show the full solution

   Substitute that \( x \) value back into the equation

6. Find the vertex of \( y = x^2 + 8x + 12 \).
   Show the full solution

   \( a = 1 \), \( b = 8 \), so the axis is \( x = -\dfrac{8}{2} = -4 \). Substituting: \( y = 16 - 32 + 12 = -4 \). The vertex is \( (-4, -4) \). Check with points on each side: at \( x = -3 \), \( y = 9 - 24 + 12 = -3 \); at \( x = -5 \), \( y = 25 - 40 + 12 = -3 \). Both are higher than \( -4 \) and equal to each other, confirming both the vertex and the symmetry. \( (-4, -4) \)

7. Find the vertex and maximum of \( y = -2x^2 + 12x - 7 \).
   Show the full solution

   \( a = -2 \), \( b = 12 \), so the axis is \( x = -\dfrac{12}{2(-2)} = -\dfrac{12}{-4} = 3 \). Substituting: \( y = -2(9) + 36 - 7 = -18 + 29 = 11 \). The vertex is \( (3, 11) \), and since \( a \) is negative this is the highest point, so the maximum value is 11. Check at \( x = 2 \): \( -8 + 24 - 7 = 9 \), which is below 11. Correct. Vertex \( (3, 11) \), maximum value 11

8. Find all intercepts of \( y = x^2 - 2x - 8 \).
   Show the full solution

   The \( y \) intercept is the constant term: \( (0, -8) \). For the \( x \) intercepts, set \( y = 0 \) and factor: two numbers multiplying to \( -8 \) and adding to \( -2 \) are \( -4 \) and 2, so \( (x - 4)(x + 2) = 0 \) and \( x = 4 \) or \( x = -2 \). The \( x \) intercepts are \( (4, 0) \) and \( (-2, 0) \). Check the symmetry: the midpoint of 4 and \( -2 \) is 1, and the vertex formula gives \( x = -\dfrac{-2}{2} = 1 \). They agree. \( (0, -8) \), \( (4, 0) \) and \( (-2, 0) \)

9. A student finds the axis of \( y = x^2 - 10x + 21 \) is \( x = 5 \) and reports the vertex as \( (5, 0) \). Correct them.
   Show the full solution

   The axis is right: \( -\dfrac{-10}{2} = 5 \). The mistake is reporting the height as 0 without substituting. The formula gives only the \( x \) coordinate of the vertex, and the \( y \) coordinate must be computed. Substituting: \( y = 25 - 50 + 21 = -4 \). The vertex is \( (5, -4) \). A quick sanity check would have caught it too. The \( y \) intercept is 21, well above the axis, and the parabola opens upward, so if the vertex were at height 0 the graph would touch the axis once. In fact the trinomial factors as \( (x-3)(x-7) \), giving two distinct intercepts at 3 and 7, which can only happen if the vertex is below the axis. \( (5, -4) \); the formula gives only \( x \), and \( y \) must be found by substituting

10. Convert \( y = x^2 + 10x + 18 \) to vertex form by completing the square, and confirm with the vertex formula.
    Show the full solution

    The coefficient of \( x \) is 10. Half of it is 5, and \( 5^2 = 25 \). Add and subtract 25 so the value is unchanged: \( y = x^2 + 10x + 25 - 25 + 18 \). The first three terms are a perfect square trinomial, from lesson 7.3: \( y = (x + 5)^2 - 7 \). So the vertex is \( (-5, -7) \). Confirming with the formula: \( a = 1 \), \( b = 10 \), so \( x = -\dfrac{10}{2} = -5 \), and substituting gives \( 25 - 50 + 18 = -7 \). The two methods agree exactly, which they must, since the formula was derived from completing the square in the first place. Checking the conversion at a third value, \( x = 0 \): standard form gives 18, and vertex form gives \( 25 - 7 = 18 \). Correct. \( y = (x + 5)^2 - 7 \), vertex \( (-5, -7) \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the zeros of \( y = (x - 4)(x - 7) \).
   Show the full solution

   4 and 7

2. Find the zeros of \( y = (x + 3)(x - 5) \).
   Show the full solution

   \( -3 \) and 5

3. Find the axis of symmetry of \( y = (x - 2)(x - 8) \).
   Show the full solution

   Midpoint of 2 and 8. \( x = 5 \)

4. What property justifies reading zeros from factors?
   Show the full solution

   The zero product property

5. Find the \( y \) intercept of \( y = (x - 1)(x - 4) \).
   Show the full solution

   At \( x = 0 \): \( (-1)(-4) = 4 \). \( (0, 4) \)

6. Find the vertex of \( y = (x + 1)(x - 7) \).
   Show the full solution

   The zeros are \( -1 \) and 7, so the axis is \( x = \dfrac{-1 + 7}{2} = 3 \). Substituting: \( y = (3 + 1)(3 - 7) = (4)(-4) = -16 \). The vertex is \( (3, -16) \). Cross-check by expanding: \( x^2 - 6x - 7 \), and the vertex formula gives \( x = 3 \) with \( y = 9 - 18 - 7 = -16 \). They agree. \( (3, -16) \)

7. Find the vertex of \( y = -2(x - 1)(x - 5) \).
   Show the full solution

   The zeros are 1 and 5, so the axis is \( x = 3 \). Substituting: \( y = -2(3 - 1)(3 - 5) = -2(2)(-2) = 8 \). The vertex is \( (3, 8) \), and since \( a = -2 \) is negative this is a maximum. Note that \( a \) changed the height of the vertex but not the zeros: the factor \( -2 \) cannot make a nonzero product zero, so it has no effect on where the graph crosses. \( (3, 8) \), a maximum

8. Write the equation with zeros \( -2 \) and 4 passing through \( (0, -16) \).
   Show the full solution

   The zeros give the brackets: \( y = a(x + 2)(x - 4) \). Substituting \( (0, -16) \): \( -16 = a(2)(-4) = -8a \), so \( a = 2 \). The equation is \( y = 2(x + 2)(x - 4) \). Check: at \( x = -2 \) and \( x = 4 \) the value is 0, and at \( x = 0 \) it is \( 2(2)(-4) = -16 \). All correct. \( y = 2(x + 2)(x - 4) \)

9. Explain why the vertex always lies halfway between the two zeros.
   Show the full solution

   A parabola is symmetric about a vertical line through its vertex, which means that for any height, the two points at that height are equally far from the axis on opposite sides. The two zeros are both at height 0, so they are such a pair. Being equally far from the axis on opposite sides means the axis is exactly at their midpoint, and the vertex lies on the axis. Algebraically, the factored form \( y = a(x - p)(x - q) \) expands to \( a\left(x^2 - (p+q)x + pq\right) \), so \( b = -a(p+q) \) and the vertex formula gives \( x = -\dfrac{-a(p+q)}{2a} = \dfrac{p+q}{2} \), which is the midpoint. This is why factored form alone is enough to sketch a graph: the zeros give the crossings, their midpoint gives the axis, and one substitution gives the vertex. Both zeros are at the same height, so symmetry puts the axis at their midpoint, and the algebra confirms \( -\frac{b}{2a} = \frac{p+q}{2} \)

10. A parabola has a single \( x \) intercept at 4 and passes through \( (6, 12) \). Find its equation and explain the single intercept.
    Show the full solution

    A single \( x \) intercept means the two zeros coincide, so both brackets are the same: \( y = a(x - 4)(x - 4) = a(x - 4)^2 \). Substituting \( (6, 12) \): \( 12 = a(2)^2 = 4a \), so \( a = 3 \). The equation is \( y = 3(x - 4)^2 \). Check: at \( x = 4 \) the value is 0, and at \( x = 6 \) it is \( 3(4) = 12 \). Both correct. The geometry behind a single intercept is that the vertex sits exactly on the \( x \) axis. The graph comes down to the axis, touches it, and goes back up without crossing. In vertex form that is \( k = 0 \), matching the condition from lesson 8.2, and in factored form it appears as a repeated factor. The two descriptions agree, since \( y = 3(x-4)^2 \) is simultaneously in both forms. \( y = 3(x - 4)^2 \); the two zeros coincide, so the vertex sits on the \( x \) axis

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Convert \( y = (x - 2)^2 + 3 \) to standard form.
   Show the full solution

   \( x^2 - 4x + 4 + 3 \). \( y = x^2 - 4x + 7 \)

2. Convert \( y = (x + 1)(x - 5) \) to standard form.
   Show the full solution

   \( y = x^2 - 4x - 5 \)

3. Convert \( y = x^2 + 5x + 6 \) to factored form.
   Show the full solution

   \( y = (x + 2)(x + 3) \)

4. Which form shows the vertex?
   Show the full solution

   Vertex form

5. Which form shows the \( y \) intercept most directly?
   Show the full solution

   Standard form, as the constant term

6. Convert \( y = x^2 - 6x + 2 \) to vertex form.
   Show the full solution

   Half of \( -6 \) is \( -3 \), and \( (-3)^2 = 9 \). Add and subtract 9: \( y = x^2 - 6x + 9 - 9 + 2 = (x - 3)^2 - 7 \). Check at \( x = 0 \): \( 9 - 7 = 2 \), matching the original constant. Correct. The vertex is \( (3, -7) \). \( y = (x - 3)^2 - 7 \)

7. Convert \( y = 3x^2 - 12x + 1 \) to vertex form.
   Show the full solution

   Factor 3 out of the first two terms: \( y = 3(x^2 - 4x) + 1 \). Half of \( -4 \) is \( -2 \), and \( (-2)^2 = 4 \): \( y = 3(x^2 - 4x + 4 - 4) + 1 \). Bringing out the \( -4 \) multiplies it by 3: \( y = 3(x - 2)^2 - 12 + 1 = 3(x - 2)^2 - 11 \). Check at \( x = 0 \): \( 3(4) - 11 = 1 \), matching. Correct. \( y = 3(x - 2)^2 - 11 \)

8. Convert \( y = 2(x + 4)^2 - 6 \) to standard form and then to factored form.
   Show the full solution

   Expanding: \( (x + 4)^2 = x^2 + 8x + 16 \), so \( y = 2x^2 + 16x + 32 - 6 = 2x^2 + 16x + 26 \). Check at \( x = 0 \): \( 2(16) - 6 = 26 \). Correct. For factored form, first take out the GCF 2: \( y = 2(x^2 + 8x + 13) \). Now look for two integers multiplying to 13 and adding to 8. Since 13 is prime, the only pair is 1 and 13, summing to 14. No integer factorization exists. So the quadratic does not factor over the integers, and standard form is as far as this conversion goes. That is a legitimate outcome: not every parabola has rational zeros, and unit 9 provides the tools for this case. \( y = 2x^2 + 16x + 26 \); it does not factor over the integers

9. A student converts \( y = 2x^2 + 8x + 3 \) and writes \( y = 2(x + 2)^2 - 1 \). Check and correct.
   Show the full solution

   Test at \( x = 0 \): the original gives 3, and their answer gives \( 2(4) - 1 = 7 \). They do not match, so the conversion is wrong. Redoing it: factor 2 out of the first two terms, \( y = 2(x^2 + 4x) + 3 \). Half of 4 is 2, and \( 2^2 = 4 \), so \( y = 2(x^2 + 4x + 4 - 4) + 3 \). Bringing out the \( -4 \) multiplies it by 2, giving \( -8 \): \( y = 2(x + 2)^2 - 8 + 3 = 2(x + 2)^2 - 5 \). Check at \( x = 0 \): \( 2(4) - 5 = 3 \). Correct. Their error was taking the \( -4 \) out of the bracket unchanged instead of multiplying it by the 2 outside. This is exactly the slip the method warns about, and the substitution check exposes it in seconds. \( y = 2(x + 2)^2 - 5 \); the subtracted constant must be multiplied by \( a \) when it leaves the bracket

10. For \( y = x^2 - 8x + 12 \), produce all three forms and say what each reveals.
    Show the full solution

    Standard form is given: \( y = x^2 - 8x + 12 \). It shows the \( y \) intercept \( (0, 12) \) directly as the constant term, and shows that the parabola opens upward since \( a = 1 \) is positive. Factored form: two numbers multiplying to 12 and adding to \( -8 \) are \( -2 \) and \( -6 \), so \( y = (x - 2)(x - 6) \). This shows the zeros at 2 and 6, so the graph crosses the \( x \) axis at \( (2, 0) \) and \( (6, 0) \). Vertex form: half of \( -8 \) is \( -4 \), and \( (-4)^2 = 16 \), so \( y = x^2 - 8x + 16 - 16 + 12 = (x - 4)^2 - 4 \). This shows the vertex \( (4, -4) \), so the minimum value of the function is \( -4 \) at \( x = 4 \). Every form checks against the others. The midpoint of the zeros 2 and 6 is 4, matching the vertex. All three evaluated at \( x = 0 \) give 12: \( 12 \), \( (-2)(-6) = 12 \), and \( 16 - 4 = 12 \). Standard \( x^2 - 8x + 12 \) shows the \( y \) intercept 12; factored \( (x-2)(x-6) \) shows the zeros 2 and 6; vertex \( (x-4)^2 - 4 \) shows the minimum \( -4 \) at \( x = 4 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What do constant first differences indicate?
   Show the full solution

   A linear function

2. What do constant second differences indicate?
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   A quadratic function

3. Classify: 2, 5, 8, 11, 14.
   Show the full solution

   First differences are all 3. Linear

4. Classify: 1, 2, 4, 8, 16.
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   Ratios are all 2. Exponential

5. Classify: 0, 1, 4, 9, 16.
   Show the full solution

   First differences 1, 3, 5, 7; second differences all 2. Quadratic

6. Classify \( x \): 0, 1, 2, 3, 4 with \( y \): 5, 9, 17, 29, 45, and find its rule.
   Show the full solution

   First differences: 4, 8, 12, 16. Not constant. Second differences: 4, 4, 4. Constant, so quadratic. The second difference is \( 2a = 4 \), so \( a = 2 \). The value at \( x = 0 \) is 5, so \( c = 5 \). The first difference from \( x = 0 \) to \( x = 1 \) is \( a + b = 4 \), so \( b = 2 \). The rule is \( y = 2x^2 + 2x + 5 \). Check at \( x = 3 \): \( 18 + 6 + 5 = 29 \). Correct. At \( x = 4 \): \( 32 + 8 + 5 = 45 \). Correct. Quadratic, \( y = 2x^2 + 2x + 5 \)

7. Classify \( x \): 0, 1, 2, 3 with \( y \): 100, 80, 64, 51.2.
   Show the full solution

   First differences: \( -20 \), \( -16 \), \( -12.8 \). Not constant. Second differences: 4, 3.2. Not constant, so not quadratic. Ratios: \( 80 \div 100 = 0.8 \), \( 64 \div 80 = 0.8 \), \( 51.2 \div 64 = 0.8 \). Constant. So it is exponential decay with \( y = 100(0.8)^x \), a 20 percent decrease per step. This is exactly the case the method warns about: the first differences were changing, which might tempt a guess of quadratic, and only the second-difference test ruled it out. Exponential, \( y = 100(0.8)^x \)

8. Is the area of a square as its side grows by 1 each time linear, quadratic or exponential?
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   The area of a square with side \( s \) is \( s^2 \), which is quadratic by definition. Confirming from a table with sides 1 through 5: areas 1, 4, 9, 16, 25. First differences: 3, 5, 7, 9. Not constant. Second differences: 2, 2, 2. Constant. So the relationship is quadratic. This is the geometric reason second differences show up: each time the side grows by 1, the area gains two strips plus a corner square, and the strips get longer by exactly one unit each time, which makes the growth of the growth constant. Quadratic

9. A student sees rising differences and concludes the data is quadratic. Explain why that is not enough.
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   Rising first differences only rule out linear. Both quadratic and exponential growth have differences that increase, so the observation does not distinguish them. What separates them is the second round. For a quadratic, the increase in the difference is itself constant: the values 1, 4, 9, 16 have differences 3, 5, 7, whose differences are 2, 2. For an exponential, the increase in the difference keeps growing: the values 1, 2, 4, 8, 16 have differences 1, 2, 4, 8, whose differences are 1, 2, 4, still rising. So the correct procedure is to take second differences, and if those are not constant, test ratios. Concluding quadratic from first differences alone will misclassify every exponential data set. Exponential data also has rising differences; only constant second differences confirm a quadratic

10. Compare \( y = x^2 \) and \( y = 2^x \) at \( x = 2, 4, 5, 10, 20 \) and describe what happens.
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    At \( x = 2 \): \( 4 \) and \( 4 \). Equal. At \( x = 4 \): \( 16 \) and \( 16 \). Equal again. At \( x = 5 \): \( 25 \) and \( 32 \). The exponential is ahead. At \( x = 10 \): \( 100 \) and \( 1024 \). About ten times ahead. At \( x = 20 \): \( 400 \) and \( 1{,}048{,}576 \). More than two thousand times ahead. Between \( x = 2 \) and \( x = 4 \) the quadratic is actually larger, for instance \( 9 \) against \( 8 \) at \( x = 3 \), which is why a short window can mislead. From \( x = 4 \) onward the exponential leads permanently and the gap widens without limit. The general principle: an exponential with base greater than 1 eventually exceeds any polynomial, however large the polynomial's degree or coefficients. Doubling applies to a number that is already doubling, while squaring applies to an input that grows only one step at a time. They tie at 2 and 4, the quadratic leads briefly in between, and from \( x = 4 \) on the exponential pulls permanently and dramatically ahead

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( h = -16t^2 + 32t \), when is the peak?
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   \( t = -\frac{32}{-32} = 1 \). At 1 second

2. What is that peak height?
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   \( -16 + 32 = 16 \). 16 feet

3. If a rectangle's perimeter is 40, and the width is \( w \), what is the length?
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   \( 2w + 2L = 40 \), so \( L = 20 - w \). \( 20 - w \)

4. Which vertex coordinate answers "how high"?
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   The \( y \) coordinate

5. Why is the domain restricted in area problems?
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   Lengths cannot be zero or negative

6. A rectangle has perimeter 40 m. Find the dimensions giving the largest area.
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   Let \( w \) be the width in meters. From \( 2w + 2L = 40 \), the length is \( L = 20 - w \). Area: \( A = w(20 - w) = -w^2 + 20w \). The leading coefficient is negative, so there is a maximum at \( w = -\dfrac{20}{2(-1)} = 10 \). Then \( L = 20 - 10 = 10 \), and \( A = 100 \) square meters. Domain: \( 0 \lt w \lt 20 \), and 10 is inside it. Check nearby: \( w = 9 \) gives \( 9 \times 11 = 99 \), and \( w = 11 \) gives \( 11 \times 9 = 99 \). Both below 100. The answer is a square, which is the general result for a fixed perimeter: among all rectangles with the same perimeter, the square has the largest area. 10 m by 10 m, area 100 square meters

7. A ball is thrown with \( h = -16t^2 + 48t + 4 \). Find the peak height and the height at 1 second.
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   Peak time: \( t = -\dfrac{48}{2(-16)} = \dfrac{48}{32} = 1.5 \) seconds. Peak height: \( h = -16(2.25) + 48(1.5) + 4 = -36 + 72 + 4 = 40 \) feet. At 1 second: \( h = -16 + 48 + 4 = 36 \) feet. The check that these are consistent: 36 is less than 40, as it must be since 1 second is before the peak. By symmetry the height at \( t = 2 \) should also be 36, and indeed \( -64 + 96 + 4 = 36 \). Peak 40 feet at 1.5 seconds; 36 feet at 1 second

8. A shop sells 200 items a week at $15. Each $1 price rise loses 10 sales. Find the price maximizing revenue.
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   Let \( x \) be the number of dollar increases. Price: \( 15 + x \) dollars. Quantity: \( 200 - 10x \) items. Revenue: \( R = (15 + x)(200 - 10x) \). Expanding: \( R = 3000 - 150x + 200x - 10x^2 = -10x^2 + 50x + 3000 \). The leading coefficient is negative, so the maximum is at \( x = -\dfrac{50}{2(-10)} = 2.5 \). Since a price can be set in half dollars, \( x = 2.5 \) is acceptable, giving a price of $17.50 and a quantity of \( 200 - 25 = 175 \) items. Revenue: \( 17.50 \times 175 = 3062.50 \) dollars. Check with the function: \( -10(6.25) + 125 + 3000 = -62.5 + 3125 = 3062.50 \). Agrees. Compare with the original: \( 15 \times 200 = 3000 \), so the rise adds $62.50 a week. $17.50, selling 175 items for $3,062.50

9. A student finds the vertex of a projectile model at \( (3, 150) \) and answers "the ball reaches its peak at 150 seconds". Correct them.
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   They have swapped the coordinates. In a height-against-time model the input is time and the output is height, so the vertex \( (3, 150) \) means the peak occurs at \( t = 3 \) seconds and the peak height is 150 feet. Their reading also fails a common-sense test: a thrown ball that stayed airborne for 150 seconds would have to reach a height of many miles, which no ordinary throw does. Checking the units of each axis before reporting, and asking whether the number is physically plausible, catches this kind of error without any algebra. The peak is at 3 seconds and the height there is 150 feet

10. A rectangular garden against a wall uses 24 m of fencing on three sides. Find the maximum area, the domain, and explain why the answer is not a square.
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    Let \( w \) be each of the two sides perpendicular to the wall, in meters. The fencing gives \( 2w + L = 24 \), so \( L = 24 - 2w \). Area: \( A = w(24 - 2w) = -2w^2 + 24w \). Maximum at \( w = -\dfrac{24}{2(-2)} = 6 \) meters, giving \( L = 24 - 12 = 12 \) meters and \( A = 6 \times 12 = 72 \) square meters. Check: \( -2(36) + 144 = 72 \). Agrees. Nearby, \( w = 5 \) gives \( 5 \times 14 = 70 \) and \( w = 7 \) gives \( 7 \times 10 = 70 \), both below 72. Domain: both dimensions must be positive, so \( w \gt 0 \) and \( 24 - 2w \gt 0 \), giving \( 0 \lt w \lt 12 \). The answer is not a square because the wall changes the cost structure. Only three sides consume fencing, so the side parallel to the wall is effectively free to extend, and the optimum puts half the fencing into that one long side and splits the other half between the two short ones. The square is optimal only when all four sides draw on the same fixed total, as in the earlier four-sided problem. 72 square meters at 6 m by 12 m, with \( 0 \lt w \lt 12 \); the free wall side makes a 1 by 2 rectangle optimal rather than a square

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( (x - 3)(x + 7) = 0 \).
   Show the full solution

   \( x = 3 \) or \( x = -7 \)

2. Solve \( x^2 - 9 = 0 \) by factoring.
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   \( (x+3)(x-3) = 0 \). \( x = \pm 3 \)

3. Solve \( x^2 + 5x + 4 = 0 \).
   Show the full solution

   \( (x+1)(x+4) = 0 \). \( x = -1 \) or \( x = -4 \)

4. Solve \( x^2 - 7x = 0 \).
   Show the full solution

   \( x(x - 7) = 0 \). The factor \( x \) gives a genuine solution. \( x = 0 \) or \( x = 7 \)

5. Why must the equation equal zero before factoring?
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   The zero product property applies only to a product of zero

6. Solve \( x^2 = 6x - 8 \).
   Show the full solution

   Rearrange to zero: \( x^2 - 6x + 8 = 0 \). Two numbers multiplying to 8 and adding to \( -6 \) are \( -2 \) and \( -4 \): \( (x - 2)(x - 4) = 0 \), so \( x = 2 \) or \( x = 4 \). Check in the original: at \( x = 2 \), \( 4 = 12 - 8 \). Correct. At \( x = 4 \), \( 16 = 24 - 8 \). Correct. \( x = 2 \) or \( x = 4 \)

7. Solve \( 2x^2 + 7x + 3 = 0 \).
   Show the full solution

   Using the method of lesson 7.6: \( ac = 6 \), and 1 and 6 add to 7. \( 2x^2 + x + 6x + 3 = x(2x+1) + 3(2x+1) = (2x+1)(x+3) = 0 \). So \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \), and \( x + 3 = 0 \) gives \( x = -3 \). Check: at \( x = -\frac{1}{2} \), \( 2\left(\frac{1}{4}\right) - \frac{7}{2} + 3 = \frac{1}{2} - \frac{7}{2} + 3 = 0 \). Correct. \( x = -\frac{1}{2} \) or \( x = -3 \)

8. Solve \( 3x^2 - 12x = 0 \).
   Show the full solution

   Factor out the GCF: \( 3x(x - 4) = 0 \). The constant 3 can never be zero, so it contributes no solution. The other factors give \( x = 0 \) and \( x = 4 \). Check: at \( x = 0 \), \( 0 - 0 = 0 \). At \( x = 4 \), \( 48 - 48 = 0 \). Both correct. A common error is dividing both sides by \( x \) at the start, which gives \( 3x = 12 \) and loses the solution \( x = 0 \) entirely. Dividing by a variable can destroy a solution, so factor instead. \( x = 0 \) or \( x = 4 \)

9. A student solves \( (x - 1)(x + 4) = 6 \) by setting \( x - 1 = 6 \) and \( x + 4 = 6 \). Explain the error and solve correctly.
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   They applied the zero product property to a product of 6, where it does not hold. A product equals 6 in infinitely many ways, so knowing the product tells you nothing about either factor individually. Their answers \( x = 7 \) and \( x = 2 \) fail: \( (6)(11) = 66 \) and \( (1)(6) = 6 \). The second happens to work by coincidence, which makes the method look plausible and is exactly why it is dangerous. Solving correctly means expanding and setting to zero: \( x^2 + 3x - 4 = 6 \), so \( x^2 + 3x - 10 = 0 \), which factors as \( (x + 5)(x - 2) = 0 \), giving \( x = -5 \) or \( x = 2 \). Check: at \( x = -5 \), \( (-6)(-1) = 6 \). Correct. At \( x = 2 \), \( (1)(6) = 6 \). Correct. \( x = -5 \) or \( x = 2 \); the property applies only against zero

10. Explain the connection between solving \( x^2 - 4x - 5 = 0 \), factoring \( x^2 - 4x - 5 \), and graphing \( y = x^2 - 4x - 5 \).
    Show the full solution

    All three are the same information in different clothing. Factoring gives \( x^2 - 4x - 5 = (x - 5)(x + 1) \), since \( -5 \) and 1 multiply to \( -5 \) and add to \( -4 \). Solving the equation sets that product to zero, so \( x = 5 \) or \( x = -1 \). Graphing \( y = x^2 - 4x - 5 \) produces a parabola crossing the horizontal axis at exactly those two points, \( (5, 0) \) and \( (-1, 0) \), because crossing the axis means \( y = 0 \), which is the equation being solved. The vocabulary reflects the three views: they are called factors, roots or solutions, and \( x \) intercepts respectively, but they are one set of numbers. This is why factoring practice in unit 7 pays off directly here, and why a graph can confirm an algebraic answer: the vertex lies at \( x = 2 \), halfway between 5 and \( -1 \), with value \( 4 - 8 - 5 = -9 \), and a parabola opening upward from \( (2, -9) \) must cross twice. The factors give the roots, the roots are the solutions, and the solutions are the \( x \) intercepts: \( x = 5 \) and \( x = -1 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( x^2 = 49 \).
   Show the full solution

   \( x = \pm 7 \)

2. Solve \( x^2 - 16 = 0 \).
   Show the full solution

   \( x = \pm 4 \)

3. Solve \( (x - 2)^2 = 9 \).
   Show the full solution

   \( x - 2 = \pm 3 \). \( x = 5 \) or \( x = -1 \)

4. Simplify \( \sqrt{50} \).
   Show the full solution

   \( 50 = 25 \times 2 \). \( 5\sqrt{2} \)

5. Does \( x^2 = -4 \) have a real solution?
   Show the full solution

   No; no real number squares to a negative

6. Solve \( 4(x + 3)^2 = 100 \).
   Show the full solution

   Divide by 4: \( (x + 3)^2 = 25 \). Take both roots: \( x + 3 = \pm 5 \), so \( x = 2 \) or \( x = -8 \). Check: \( 4(2 + 3)^2 = 4(25) = 100 \). Correct. \( 4(-8 + 3)^2 = 4(25) = 100 \). Correct. \( x = 2 \) or \( x = -8 \)

7. Solve \( (x - 5)^2 = 48 \), leaving the answer exact.
   Show the full solution

   Take both roots: \( x - 5 = \pm\sqrt{48} \). Simplify: \( 48 = 16 \times 3 \), so \( \sqrt{48} = 4\sqrt{3} \). So \( x = 5 \pm 4\sqrt{3} \). Check the positive case: \( x - 5 = 4\sqrt{3} \), and squaring gives \( 16 \times 3 = 48 \). Correct. \( x = 5 \pm 4\sqrt{3} \)

8. Solve \( 2x^2 + 5 = 55 \).
   Show the full solution

   Subtract 5: \( 2x^2 = 50 \). Divide by 2: \( x^2 = 25 \). Take both roots: \( x = \pm 5 \). Check: \( 2(25) + 5 = 55 \), and the same for \( x = -5 \). Correct. Note the order of operations in reverse: the added 5 came off first, then the multiplication by 2, and only then the square. Dividing by 2 before subtracting would have given \( x^2 + 2.5 = 27.5 \), which is still correct but less direct. \( x = \pm 5 \)

9. A student solves \( (x - 3)^2 = 16 \) and answers \( x = 7 \). Explain what is missing.
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   They took only the positive square root. Both 4 and \( -4 \) square to 16, so the equation \( (x - 3)^2 = 16 \) gives two cases: \( x - 3 = 4 \), giving \( x = 7 \), and \( x - 3 = -4 \), giving \( x = -1 \). Checking the missing one: \( (-1 - 3)^2 = (-4)^2 = 16 \). Correct, so it is a genuine solution and not an artifact. Graphically the reason is clear: \( y = (x - 3)^2 \) is a parabola with vertex \( (3, 0) \), and a horizontal line at height 16 cuts it twice, once on each side of the axis of symmetry. A quadratic equation with a positive isolated square always has two solutions, symmetric about the vertex. The second solution \( x = -1 \); both square roots must be taken

10. Solve \( 3(2x - 1)^2 - 27 = 0 \) and explain why the inner expression makes no difference to the method.
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    Add 27: \( 3(2x - 1)^2 = 27 \). Divide by 3: \( (2x - 1)^2 = 9 \). Take both roots: \( 2x - 1 = \pm 3 \). Case one: \( 2x - 1 = 3 \), so \( 2x = 4 \) and \( x = 2 \). Case two: \( 2x - 1 = -3 \), so \( 2x = -2 \) and \( x = -1 \). Check: \( 3(2(2) - 1)^2 - 27 = 3(9) - 27 = 0 \). Correct. \( 3(2(-1) - 1)^2 - 27 = 3(-3)^2 - 27 = 0 \). Correct. The inner expression makes no difference because the square root step treats whatever is being squared as a single object. Whether it is \( x \), \( x - 4 \) or \( 2x - 1 \), the step says that object equals plus or minus the root. Only afterward does the specific expression matter, and then it is an ordinary linear equation of the kind solved in unit 2. That is what makes this method short: it converts one quadratic into two linear problems. \( x = 2 \) or \( x = -1 \); the squared expression is treated as a single object until the roots are taken

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What number completes the square for \( x^2 + 8x \)?
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   Half of 8 is 4, squared is 16. 16

2. What number completes the square for \( x^2 - 10x \)?
   Show the full solution

   25

3. Factor \( x^2 + 12x + 36 \).
   Show the full solution

   \( (x + 6)^2 \)

4. Solve \( x^2 + 4x = 5 \) by completing the square.
   Show the full solution

   Add 4: \( x^2 + 4x + 4 = 9 \), so \( (x+2)^2 = 9 \) and \( x + 2 = \pm 3 \). \( x = 1 \) or \( x = -5 \)

5. Why must the completing number be added to both sides?
   Show the full solution

   To keep the equation balanced and equivalent

6. Solve \( x^2 - 6x - 16 = 0 \) by completing the square.
   Show the full solution

   Move the constant: \( x^2 - 6x = 16 \). Half of \( -6 \) is \( -3 \), squared is 9. Add 9 to both sides: \( x^2 - 6x + 9 = 25 \), so \( (x - 3)^2 = 25 \). Take both roots: \( x - 3 = \pm 5 \), giving \( x = 8 \) or \( x = -2 \). Check: \( 64 - 48 - 16 = 0 \) and \( 4 + 12 - 16 = 0 \). Both correct. \( x = 8 \) or \( x = -2 \)

7. Solve \( x^2 + 10x + 7 = 0 \), leaving the answer exact.
   Show the full solution

   Move the constant: \( x^2 + 10x = -7 \). Half of 10 is 5, squared is 25. Add 25 to both sides: \( x^2 + 10x + 25 = 18 \), so \( (x + 5)^2 = 18 \). Take roots: \( x + 5 = \pm\sqrt{18} \). Since \( 18 = 9 \times 2 \), \( \sqrt{18} = 3\sqrt{2} \). So \( x = -5 \pm 3\sqrt{2} \). Check the structure: \( (x+5)^2 = (\pm 3\sqrt{2})^2 = 9 \times 2 = 18 \). Correct. \( x = -5 \pm 3\sqrt{2} \)

8. Solve \( 3x^2 + 12x - 15 = 0 \).
   Show the full solution

   Divide every term by 3 first: \( x^2 + 4x - 5 = 0 \), so \( x^2 + 4x = 5 \). Half of 4 is 2, squared is 4. Add 4 to both sides: \( x^2 + 4x + 4 = 9 \), so \( (x + 2)^2 = 9 \) and \( x + 2 = \pm 3 \). That gives \( x = 1 \) or \( x = -5 \). Check in the original: \( 3 + 12 - 15 = 0 \) and \( 75 - 60 - 15 = 0 \). Both correct. Dividing by the leading coefficient first is essential; completing the square on \( 3x^2 + 12x \) directly does not work, because the method is built for a leading coefficient of 1. \( x = 1 \) or \( x = -5 \)

9. Explain the geometric picture that gives the method its name.
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   Think of \( x^2 + bx \) as area. The \( x^2 \) is a square of side \( x \). The \( bx \) can be drawn as a rectangle of dimensions \( b \) by \( x \), and splitting it in half gives two strips of width \( \frac{b}{2} \) and length \( x \). Attach one strip to the right edge of the square and the other to the bottom edge. The figure is now almost a larger square of side \( x + \frac{b}{2} \), but a small corner is missing where the two strips would meet. That corner is a square of side \( \frac{b}{2} \), so its area is \( \left(\frac{b}{2}\right)^2 \). Adding exactly that amount fills the gap and completes the square, and the resulting figure has area \( \left(x + \frac{b}{2}\right)^2 \). That is precisely the algebraic identity the method uses, and it explains both the name and why half of \( b \), not \( b \) itself, is what gets squared. The two half-strips leave a corner gap of \( \left(\frac{b}{2}\right)^2 \), and filling it completes a square of side \( x + \frac{b}{2} \)

10. Solve \( 2x^2 - 12x + 5 = 0 \) exactly, and say why factoring was not an option.
    Show the full solution

    Divide every term by 2: \( x^2 - 6x + 2.5 = 0 \), so \( x^2 - 6x = -2.5 \). Half of \( -6 \) is \( -3 \), squared is 9. Add 9 to both sides: \( x^2 - 6x + 9 = 6.5 \), so \( (x - 3)^2 = 6.5 \). Take roots: \( x - 3 = \pm\sqrt{6.5} \). Writing 6.5 as \( \frac{13}{2} \): \( \sqrt{\frac{13}{2}} = \frac{\sqrt{13}}{\sqrt{2}} = \frac{\sqrt{26}}{2} \). So \( x = 3 \pm \dfrac{\sqrt{26}}{2} \), roughly 5.550 and 0.450. Check numerically: \( 2(5.550)^2 - 12(5.550) + 5 \approx 61.60 - 66.60 + 5 \approx 0 \). Correct to rounding. Factoring was not an option because the solutions are irrational. Factoring over the integers produces rational roots, so an equation with roots involving \( \sqrt{26} \) cannot factor that way. Trying \( ac = 10 \) confirms it: the pairs 1 and 10, and 2 and 5, give sums of 11 and 7, never \( -12 \). Completing the square has no such restriction, which is why it solves every quadratic. \( x = 3 \pm \frac{\sqrt{26}}{2} \); the roots are irrational, so no integer factorization exists

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. For \( x^2 + 5x + 6 = 0 \), what are \( a \), \( b \) and \( c \)?
   Show the full solution

   \( a = 1 \), \( b = 5 \), \( c = 6 \)

2. Compute \( b^2 - 4ac \) for that equation.
   Show the full solution

   \( 25 - 24 \). 1

3. Finish solving it.
   Show the full solution

   \( x = \dfrac{-5 \pm 1}{2} \). \( x = -2 \) or \( x = -3 \)

4. If \( b = -4 \), what is \( -b \)?
   Show the full solution

   \( 4 \)

5. If \( b = -4 \), what is \( b^2 \)?
   Show the full solution

   16, since squaring removes the sign

6. Solve \( x^2 - 2x - 15 = 0 \) with the formula.
   Show the full solution

   \( a = 1 \), \( b = -2 \), \( c = -15 \). \( b^2 - 4ac = 4 - 4(1)(-15) = 4 + 60 = 64 \). \( x = \dfrac{-(-2) \pm \sqrt{64}}{2} = \dfrac{2 \pm 8}{2} \), giving \( x = 5 \) or \( x = -3 \). Check: \( 25 - 10 - 15 = 0 \) and \( 9 + 6 - 15 = 0 \). Both correct. Factoring would have given the same answers faster here, since \( (x-5)(x+3) \) works. \( x = 5 \) or \( x = -3 \)

7. Solve \( 3x^2 + 2x - 8 = 0 \).
   Show the full solution

   \( a = 3 \), \( b = 2 \), \( c = -8 \). \( b^2 - 4ac = 4 - 4(3)(-8) = 4 + 96 = 100 \). \( x = \dfrac{-2 \pm 10}{6} \), giving \( x = \dfrac{8}{6} = \dfrac{4}{3} \) or \( x = \dfrac{-12}{6} = -2 \). Check at \( x = -2 \): \( 12 - 4 - 8 = 0 \). Correct. \( x = \frac{4}{3} \) or \( x = -2 \)

8. Solve \( x^2 - 4x - 1 = 0 \), leaving the answer exact.
   Show the full solution

   \( a = 1 \), \( b = -4 \), \( c = -1 \). \( b^2 - 4ac = 16 - 4(1)(-1) = 16 + 4 = 20 \). \( x = \dfrac{4 \pm \sqrt{20}}{2} = \dfrac{4 \pm 2\sqrt{5}}{2} \). Every term in the numerator has a factor of 2, so it reduces to \( x = 2 \pm \sqrt{5} \). Check at \( x = 2 + \sqrt{5} \): \( (2+\sqrt{5})^2 = 4 + 4\sqrt{5} + 5 \), and subtracting \( 4(2 + \sqrt{5}) = 8 + 4\sqrt{5} \) and 1 gives \( 9 + 4\sqrt{5} - 8 - 4\sqrt{5} - 1 = 0 \). Correct. \( x = 2 \pm \sqrt{5} \)

9. A student solves \( x^2 - 6x + 5 = 0 \) and computes \( b^2 - 4ac = -36 - 20 = -56 \). Find both errors.
   Show the full solution

   Two separate mistakes. First, \( b^2 \) with \( b = -6 \) is \( (-6)^2 = 36 \), not \( -36 \); squaring a negative gives a positive, and they applied the minus outside the square. Second, \( 4ac = 4(1)(5) = 20 \), and the formula subtracts it, giving \( 36 - 20 = 16 \), not \( -36 - 20 \). With the correct value: \( x = \dfrac{6 \pm \sqrt{16}}{2} = \dfrac{6 \pm 4}{2} \), giving \( x = 5 \) or \( x = 1 \). Check: \( 25 - 30 + 5 = 0 \) and \( 1 - 6 + 5 = 0 \). Both correct. Their negative answer should have been a warning in itself: a negative radicand means no real solutions, and \( x^2 - 6x + 5 \) factors as \( (x-5)(x-1) \), so it obviously has two. \( b^2 = 36 \), not \( -36 \), and \( 4ac \) is subtracted, giving 16; the solutions are \( x = 5 \) and \( x = 1 \)

10. Solve \( 2x^2 - 4x - 3 = 0 \) exactly and as decimals, showing every simplification.
    Show the full solution

    \( a = 2 \), \( b = -4 \), \( c = -3 \). Radicand: \( b^2 - 4ac = 16 - 4(2)(-3) = 16 + 24 = 40 \). Formula: \( x = \dfrac{-(-4) \pm \sqrt{40}}{2(2)} = \dfrac{4 \pm \sqrt{40}}{4} \). Simplify the radical: \( 40 = 4 \times 10 \), so \( \sqrt{40} = 2\sqrt{10} \), giving \( x = \dfrac{4 \pm 2\sqrt{10}}{4} \). Reduce: every term in the numerator has a factor of 2, and so does the denominator, so \( x = \dfrac{2 \pm \sqrt{10}}{2} \). No further reduction is possible, since 2 does not divide \( \sqrt{10} \). Canceling the 2 in the denominator against only the 2 in the numerator would be wrong for exactly that reason. As decimals, \( \sqrt{10} \approx 3.162 \), so \( x \approx \dfrac{5.162}{2} \approx 2.581 \) or \( x \approx \dfrac{-1.162}{2} \approx -0.581 \). Check the first: \( 2(2.581)^2 - 4(2.581) - 3 \approx 13.32 - 10.32 - 3 \approx 0 \). Correct to rounding. \( x = \dfrac{2 \pm \sqrt{10}}{2} \), about 2.581 and \( -0.581 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. What is the discriminant of \( x^2 + 3x + 1 = 0 \)?
   Show the full solution

   \( 9 - 4 \). 5

2. How many real solutions does a discriminant of 25 predict?
   Show the full solution

   Two, and they are rational

3. How many for a discriminant of 0?
   Show the full solution

   Exactly one

4. How many for a discriminant of \( -8 \)?
   Show the full solution

   None

5. What does a discriminant of 0 mean about the graph?
   Show the full solution

   The vertex sits on the \( x \) axis and the parabola touches once

6. Find the discriminant of \( 2x^2 - 7x + 3 = 0 \) and describe the solutions.
   Show the full solution

   \( a = 2 \), \( b = -7 \), \( c = 3 \), so \( b^2 - 4ac = 49 - 24 = 25 \). Positive, so two real solutions, and 25 is a perfect square, so they are rational and the quadratic factors over the integers. Confirming: \( (2x - 1)(x - 3) = 0 \) gives \( x = \frac{1}{2} \) and \( x = 3 \). Both rational, as predicted. 25; two rational solutions

7. Find the discriminant of \( x^2 + 4x + 7 = 0 \) and interpret it graphically.
   Show the full solution

   \( b^2 - 4ac = 16 - 28 = -12 \). Negative, so there are no real solutions. Graphically, the vertex is at \( x = -\dfrac{4}{2} = -2 \) with \( y = 4 - 8 + 7 = 3 \), so the vertex is \( (-2, 3) \). Since \( a = 1 \) is positive, the parabola opens upward from a lowest point three units above the horizontal axis, so it never touches the axis. \( -12 \); the parabola lies entirely above the \( x \) axis, with vertex \( (-2, 3) \)

8. Find \( k \) making \( x^2 + kx + 9 = 0 \) have exactly one solution.
   Show the full solution

   Exactly one solution means the discriminant is zero: \( k^2 - 4(1)(9) = 0 \), so \( k^2 = 36 \) and \( k = \pm 6 \). Both roots must be taken, as in lesson 9.2. Checking \( k = 6 \): \( x^2 + 6x + 9 = (x+3)^2 = 0 \), giving the single solution \( x = -3 \). Checking \( k = -6 \): \( x^2 - 6x + 9 = (x-3)^2 = 0 \), giving the single solution \( x = 3 \). Both work, so there are two values of \( k \), each producing a perfect square trinomial. \( k = 6 \) or \( k = -6 \)

9. Explain why a discriminant of zero produces one solution rather than none.
   Show the full solution

   Look at the formula: \( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). With a discriminant of zero the radical becomes \( \sqrt{0} = 0 \), which is a perfectly ordinary real number. Adding and subtracting zero give the same result, so both branches collapse to \( x = -\dfrac{b}{2a} \), a single solution. Notice that \( -\dfrac{b}{2a} \) is exactly the vertex formula from lesson 8.3. That is the geometry: the one solution occurs where the vertex sits on the horizontal axis, so the parabola touches the axis at its turning point without crossing. The no-solution case is different in kind. It requires the square root of a negative, which does not exist among the real numbers, so the formula produces nothing real at all. Zero under the radical is a value; a negative under the radical is an impossibility. The radical is zero, so both branches give the same answer \( x = -\frac{b}{2a} \), which is the vertex sitting on the axis

10. For \( x^2 - 6x + k = 0 \), describe how the number of solutions changes as \( k \) varies.
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    The discriminant is \( 36 - 4k \), and its sign decides everything. Two real solutions require \( 36 - 4k \gt 0 \), so \( 4k \lt 36 \) and \( k \lt 9 \). Exactly one requires \( 36 - 4k = 0 \), so \( k = 9 \). None requires \( 36 - 4k \lt 0 \), so \( k \gt 9 \). Checking the boundary: at \( k = 9 \), the equation is \( x^2 - 6x + 9 = (x-3)^2 = 0 \), with the single solution \( x = 3 \). Checking either side: at \( k = 8 \), the discriminant is 4 and the solutions are \( x = \dfrac{6 \pm 2}{2} \), giving 4 and 2. At \( k = 10 \), the discriminant is \( -4 \) and there are none. The geometry explains it. The vertex is always at \( x = 3 \), and its height is \( 9 - 18 + k = k - 9 \). As \( k \) increases, the whole parabola rises. While the vertex is below the axis there are two crossings; when it reaches the axis at \( k = 9 \) the two crossings merge into one; above that the parabola clears the axis entirely. \( k \lt 9 \) gives two, \( k = 9 \) gives one, \( k \gt 9 \) gives none, as the vertex rises through the axis

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Which method suits \( x^2 = 81 \)?
   Show the full solution

   Square roots

2. Which method suits \( x^2 + 6x + 5 = 0 \)?
   Show the full solution

   Factoring

3. Which method always works?
   Show the full solution

   The quadratic formula

4. Solve \( x^2 - 9x = 0 \).
   Show the full solution

   \( x(x - 9) = 0 \). \( x = 0 \) or \( x = 9 \)

5. Why should you not divide both sides by \( x \)?
   Show the full solution

   It loses the solution \( x = 0 \)

6. Choose a method for \( 5x^2 - 80 = 0 \) and solve.
   Show the full solution

   There is no first-power term, so square roots is fastest. Add 80: \( 5x^2 = 80 \). Divide by 5: \( x^2 = 16 \). Take both roots: \( x = \pm 4 \). Check: \( 5(16) - 80 = 0 \) for both signs. Correct. Factoring would also have worked here, since \( 5(x^2 - 16) = 5(x+4)(x-4) \), and gives the same answers. Both are fast; the formula would not have been. \( x = \pm 4 \)

7. Choose a method for \( 2x^2 + 3x - 7 = 0 \) and solve.
   Show the full solution

   Check the discriminant first: \( b^2 - 4ac = 9 - 4(2)(-7) = 9 + 56 = 65 \). Positive but not a perfect square, so factoring cannot work and the formula is the method. \( x = \dfrac{-3 \pm \sqrt{65}}{4} \). Since 65 factors as \( 5 \times 13 \) with no perfect square factor, the radical does not simplify, and the answer is exact as written. As decimals, \( \sqrt{65} \approx 8.062 \), giving about 1.266 and \( -2.766 \). \( x = \dfrac{-3 \pm \sqrt{65}}{4} \)

8. Choose a method for \( 3x^2 - 27x + 60 = 0 \) and solve.
   Show the full solution

   Factor out the GCF 3 first, as always: \( 3(x^2 - 9x + 20) = 0 \). Since 3 is never zero, the solutions come from the bracket. Now the leading coefficient is 1 and the numbers are small, so factor: two numbers multiplying to 20 and adding to \( -9 \) are \( -4 \) and \( -5 \). \( 3(x - 4)(x - 5) = 0 \), so \( x = 4 \) or \( x = 5 \). Check: \( 48 - 108 + 60 = 0 \) and \( 75 - 135 + 60 = 0 \). Both correct. Going straight to the formula with \( a = 3 \), \( b = -27 \), \( c = 60 \) would have meant computing \( 729 - 720 = 9 \) and then \( \dfrac{27 \pm 3}{6} \), which works but handles much larger numbers. \( x = 4 \) or \( x = 5 \)

9. A student uses the quadratic formula on \( (x - 5)^2 = 36 \). Show a faster route.
   Show the full solution

   Their route works but is long: expanding gives \( x^2 - 10x + 25 = 36 \), then \( x^2 - 10x - 11 = 0 \), then identifying coefficients, computing \( 100 + 44 = 144 \), and evaluating \( \dfrac{10 \pm 12}{2} \) to get 11 and \( -1 \). The square is already isolated, so square roots finishes in two lines: \( x - 5 = \pm 6 \), giving \( x = 11 \) or \( x = -1 \). Same answers, a fraction of the work and far fewer places to make an arithmetic slip. The general rule is that an isolated square never needs expanding. Expanding a perfect square only to rebuild it with the formula undoes work that was already done. Take square roots directly: \( x - 5 = \pm 6 \), so \( x = 11 \) or \( x = -1 \)

10. Solve \( x^2 + 8x + 12 = 0 \) by three different methods and compare the effort.
    Show the full solution

    Factoring. Two numbers multiplying to 12 and adding to 8 are 2 and 6: \( (x + 2)(x + 6) = 0 \), giving \( x = -2 \) or \( x = -6 \). About fifteen seconds. Completing the square. \( x^2 + 8x = -12 \). Half of 8 is 4, squared is 16, so \( x^2 + 8x + 16 = 4 \) and \( (x + 4)^2 = 4 \), giving \( x + 4 = \pm 2 \) and the same two answers. About forty seconds, and it also reveals the vertex \( (-4, -4) \). Quadratic formula. \( b^2 - 4ac = 64 - 48 = 16 \), so \( x = \dfrac{-8 \pm 4}{2} \), giving \( -2 \) and \( -6 \). About a minute with three coefficient substitutions to get right. All three agree, as they must. Factoring is fastest here because the discriminant 16 is a perfect square and the numbers are small. Completing the square costs more but delivers the vertex as a bonus. The formula is the most reliable and the slowest, and it is the only one of the three that would still have worked had the constant been 13 instead of 12. \( x = -2 \) or \( x = -6 \) by all three; factoring is fastest, completing the square also gives the vertex, and the formula is the most general

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Two numbers differ by 3 and multiply to 40. Write the equation.
   Show the full solution

   \( x(x + 3) = 40 \)

2. Solve it for positive \( x \).
   Show the full solution

   \( x^2 + 3x - 40 = 0 \), so \( (x+8)(x-5) = 0 \). \( x = 5 \), giving 5 and 8

3. Why is a negative time usually rejected?
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   It falls before the motion started, where the model does not apply

4. For \( h = -16t^2 + 32t \), when does the object land?
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   \( -16t(t - 2) = 0 \), so \( t = 0 \) or \( t = 2 \); the first is the launch. At 2 seconds

5. A square's area is 49. What is its side?
   Show the full solution

   \( s^2 = 49 \) gives \( s = \pm 7 \), and a length cannot be negative. 7

6. A rectangle's length is 5 cm more than its width and its area is 84 square cm. Find both dimensions.
   Show the full solution

   Let \( w \) be the width in cm, so the length is \( w + 5 \). \( w(w + 5) = 84 \), so \( w^2 + 5w - 84 = 0 \). Two numbers multiplying to \( -84 \) and adding to 5 are 12 and \( -7 \): \( (w + 12)(w - 7) = 0 \), giving \( w = -12 \) or \( w = 7 \). Reject \( w = -12 \) as a negative length. The width is 7 cm and the length is 12 cm. Check: \( 12 = 7 + 5 \), and \( 7 \times 12 = 84 \). Both conditions hold. 7 cm by 12 cm

7. A ball is thrown with \( h = -16t^2 + 32t + 48 \). Find when it lands and its peak height.
   Show the full solution

   Landing means \( h = 0 \): \( -16t^2 + 32t + 48 = 0 \). Factor out \( -16 \): \( -16(t^2 - 2t - 3) = 0 \), so \( (t - 3)(t + 1) = 0 \) and \( t = 3 \) or \( t = -1 \). Reject \( t = -1 \) as before the throw. The ball lands after 3 seconds. Peak: \( t = -\dfrac{32}{2(-16)} = 1 \) second, and \( h = -16 + 32 + 48 = 64 \) feet. Checks: at \( t = 0 \) the height is 48 ft, the throwing height; the peak at 1 second lies between 0 and 3; and \( -16(9) + 96 + 48 = -144 + 144 = 0 \) confirms the landing. Lands at 3 seconds; peak height 64 feet at 1 second

8. A photo 8 cm by 10 cm is given a uniform border, making the total area 168 square cm. Find the border width.
   Show the full solution

   Let \( x \) be the border width in cm. The border adds \( x \) to each side, so it adds \( 2x \) to each dimension: the framed piece is \( 8 + 2x \) by \( 10 + 2x \). \( (8 + 2x)(10 + 2x) = 168 \). Expanding: \( 80 + 16x + 20x + 4x^2 = 168 \), so \( 4x^2 + 36x - 88 = 0 \). Divide by 4: \( x^2 + 9x - 22 = 0 \). Two numbers multiplying to \( -22 \) and adding to 9 are 11 and \( -2 \): \( (x + 11)(x - 2) = 0 \), giving \( x = -11 \) or \( x = 2 \). Reject \( x = -11 \) as a negative width. The border is 2 cm wide. Check: the framed piece is 12 by 14, with area \( 168 \) square cm. Correct. The step most often missed is the doubling: a border of width \( x \) appears on both sides of each dimension, so each grows by \( 2x \), not \( x \). 2 cm

9. A student solves an area problem and reports both \( w = 6 \) and \( w = -9 \). Explain what should happen.
   Show the full solution

   Both are genuine solutions of the equation, and the algebra is not wrong. But the variable represents a width, and a physical width cannot be negative, so \( w = -9 \) does not correspond to any rectangle. It must be discarded, and the reason must be stated rather than left implicit. The correct presentation names both solutions, rejects one with its justification, and reports the other with units: "the solutions are 6 and \( -9 \); a width cannot be negative, so the width is 6 units." Two errors to avoid on either side of this: reporting both as answers, which shows the context was not considered, and never mentioning the negative solution at all, which hides a step of the reasoning. The rejection is part of the mathematics. Keep \( w = 6 \), reject \( w = -9 \) explicitly because a width cannot be negative

10. A company's profit is \( P = -2x^2 + 80x - 350 \) dollars for \( x \) units. Find the break-even points and the maximum profit.
    Show the full solution

    Breaking even means zero profit: \( -2x^2 + 80x - 350 = 0 \). Divide by \( -2 \): \( x^2 - 40x + 175 = 0 \). The discriminant is \( 1600 - 700 = 900 \), a perfect square, so it factors. Two numbers multiplying to 175 and adding to \( -40 \) are \( -5 \) and \( -35 \): \( (x - 5)(x - 35) = 0 \), giving \( x = 5 \) or \( x = 35 \). Both are positive whole numbers of units, so both are meaningful and both are kept. The company breaks even at 5 units and again at 35 units, losing money below 5 and above 35 and profiting between them. Maximum profit is the vertex: \( x = -\dfrac{80}{2(-2)} = 20 \) units, and \( P = -2(400) + 1600 - 350 = -800 + 1250 = 450 \) dollars. Checks: 20 lies halfway between 5 and 35, as the vertex must. At \( x = 5 \), \( -50 + 400 - 350 = 0 \). Correct. At \( x = 40 \), \( -3200 + 3200 - 350 = -350 \), a loss, consistent with being past the second break-even point. This problem is the counterpart to the previous one: here neither solution is rejected, because both describe a real production level. Examining each solution against the context does not always mean discarding one. Break-even at 5 and 35 units; maximum profit $450 at 20 units

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \sqrt{36} \).
   Show the full solution

   6

2. Simplify \( \sqrt{18} \).
   Show the full solution

   \( 18 = 9 \times 2 \). \( 3\sqrt{2} \)

3. Simplify \( \sqrt{45} \).
   Show the full solution

   \( 45 = 9 \times 5 \). \( 3\sqrt{5} \)

4. Simplify \( \sqrt{x^8} \).
   Show the full solution

   Half the exponent. \( x^4 \)

5. Does \( \sqrt{4 + 9} = 2 + 3 \)?
   Show the full solution

   \( \sqrt{13} \approx 3.61 \), not 5. No

6. Simplify \( \sqrt{200} \).
   Show the full solution

   The largest perfect square factor of 200 is 100, since \( 200 = 100 \times 2 \). \( \sqrt{200} = 10\sqrt{2} \). Check by squaring: \( 100 \times 2 = 200 \). Correct. Numerically, \( \sqrt{200} \approx 14.14 \) and \( 10\sqrt{2} \approx 14.14 \). \( 10\sqrt{2} \)

7. Simplify \( \sqrt{27x^5} \).
   Show the full solution

   Number part: \( 27 = 9 \times 3 \), so 3 comes out and 3 stays. Variable part: \( x^5 = x^4 \cdot x \), and \( \sqrt{x^4} = x^2 \), so \( x^2 \) comes out and one \( x \) stays. \( \sqrt{27x^5} = 3x^2\sqrt{3x} \). Check by squaring: \( 9x^4 \cdot 3x = 27x^5 \). Correct. \( 3x^2\sqrt{3x} \)

8. Simplify \( 2\sqrt{75} \).
   Show the full solution

   \( 75 = 25 \times 3 \), so \( \sqrt{75} = 5\sqrt{3} \). The coefficient 2 multiplies what comes out: \( 2 \times 5\sqrt{3} = 10\sqrt{3} \). Check numerically: \( 2\sqrt{75} \approx 2 \times 8.660 \approx 17.32 \), and \( 10\sqrt{3} \approx 17.32 \). Correct. \( 10\sqrt{3} \)

9. Explain why \( \sqrt{16} \) is 4 and not \( \pm 4 \), while \( x^2 = 16 \) has two solutions.
   Show the full solution

   The radical symbol is defined to output a single number, because expressions must evaluate unambiguously. If \( \sqrt{16} \) meant both 4 and \( -4 \), then an expression like \( \sqrt{16} + 1 \) would have no definite value, and the quadratic formula would be meaningless since it already supplies its own plus-or-minus. The convention picks the nonnegative root, called the principal root. The equation \( x^2 = 16 \) is a different kind of question. It asks which numbers square to 16, and both 4 and \( -4 \) do, so it genuinely has two solutions. They are written \( x = \pm\sqrt{16} = \pm 4 \), with the plus-or-minus added deliberately by whoever is solving. So the symbol names one number; the equation has two answers; and the plus-or-minus sign is what bridges them. The symbol names the principal nonnegative root, while the equation asks for every number whose square is 16

10. Simplify \( \sqrt{98x^4 y^7} \), showing each part separately.
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    Number: the perfect square factors of 98 are found from \( 98 = 49 \times 2 \), so \( \sqrt{49} = 7 \) comes out and 2 stays. The \( x \) part: the exponent 4 is even, so \( \sqrt{x^4} = x^2 \) comes out entirely with nothing left behind. The \( y \) part: the exponent 7 is odd, so write \( y^7 = y^6 \cdot y \). Then \( \sqrt{y^6} = y^3 \) comes out and one \( y \) stays. Combining: \( \sqrt{98x^4y^7} = 7x^2y^3\sqrt{2y} \). Check by squaring: \( (7x^2y^3)^2 = 49x^4y^6 \), and multiplying by the remaining \( 2y \) gives \( 98x^4y^7 \). Correct. The general rule the parts illustrate: an even exponent comes out completely as half itself, and an odd exponent leaves exactly one factor behind. \( 7x^2 y^3 \sqrt{2y} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( 3\sqrt{7} + 5\sqrt{7} \).
   Show the full solution

   \( 8\sqrt{7} \)

2. Can \( \sqrt{2} + \sqrt{5} \) be combined?
   Show the full solution

   No; the radicands differ

3. Simplify \( \sqrt{6} \cdot \sqrt{10} \).
   Show the full solution

   \( \sqrt{60} = \sqrt{4 \times 15} \). \( 2\sqrt{15} \)

4. Simplify \( \dfrac{10}{\sqrt{5}} \).
   Show the full solution

   \( \dfrac{10\sqrt{5}}{5} \). \( 2\sqrt{5} \)

5. What is the conjugate of \( 2 - \sqrt{7} \)?
   Show the full solution

   \( 2 + \sqrt{7} \)

6. Simplify \( \sqrt{50} + \sqrt{8} - \sqrt{18} \).
   Show the full solution

   Simplify each first: \( \sqrt{50} = 5\sqrt{2} \); \( \sqrt{8} = 2\sqrt{2} \); \( \sqrt{18} = 3\sqrt{2} \). All three are now like radicals: \( 5\sqrt{2} + 2\sqrt{2} - 3\sqrt{2} = 4\sqrt{2} \). Check numerically: \( 7.071 + 2.828 - 4.243 \approx 5.657 \), and \( 4(1.414) \approx 5.657 \). Correct. \( 4\sqrt{2} \)

7. Expand \( (\sqrt{3} + 4)^2 \).
   Show the full solution

   Using the square pattern with \( a = \sqrt{3} \) and \( b = 4 \): \( a^2 = 3 \); \( 2ab = 2(\sqrt{3})(4) = 8\sqrt{3} \); \( b^2 = 16 \). So \( (\sqrt{3} + 4)^2 = 3 + 8\sqrt{3} + 16 = 19 + 8\sqrt{3} \). Check numerically: \( (1.732 + 4)^2 = (5.732)^2 \approx 32.86 \), and \( 19 + 8(1.732) \approx 19 + 13.86 \approx 32.86 \). Correct. The middle term is the one most often dropped, exactly as in lesson 7.3. \( 19 + 8\sqrt{3} \)

8. Rationalize \( \dfrac{6}{2 + \sqrt{3}} \).
   Show the full solution

   Multiply top and bottom by the conjugate \( 2 - \sqrt{3} \). Denominator: \( (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \). Numerator: \( 6(2 - \sqrt{3}) = 12 - 6\sqrt{3} \). So the result is \( \dfrac{12 - 6\sqrt{3}}{1} = 12 - 6\sqrt{3} \). Check numerically: \( 6 \div (2 + 1.732) = 6 \div 3.732 \approx 1.608 \), and \( 12 - 6(1.732) = 12 - 10.39 \approx 1.61 \). Correct. \( 12 - 6\sqrt{3} \)

9. A student writes \( \sqrt{9} + \sqrt{16} = \sqrt{25} \). Explain the error.
   Show the full solution

   They applied a rule that does not exist. Evaluating both sides settles it: \( \sqrt{9} + \sqrt{16} = 3 + 4 = 7 \), while \( \sqrt{25} = 5 \). The claim is simply false. The rule that does hold is for products: \( \sqrt{a}\sqrt{b} = \sqrt{ab} \), which checks out since \( \sqrt{9}\sqrt{16} = 12 \) and \( \sqrt{144} = 12 \). The reason there is no sum rule is the middle term. Squaring \( \sqrt{a} + \sqrt{b} \) gives \( a + 2\sqrt{ab} + b \), which exceeds \( a + b \) whenever both are positive. In this case squaring 7 gives 49, not 25, and the extra \( 2\sqrt{144} = 24 \) accounts for the whole difference. \( \sqrt{9} + \sqrt{16} = 7 \), not 5; radicals distribute over multiplication but never over addition

10. Simplify \( \dfrac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \) completely.
    Show the full solution

    The conjugate of \( \sqrt{2} + \sqrt{3} \) is \( \sqrt{2} - \sqrt{3} \). Multiply top and bottom by it. Denominator: \( (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = 2 - 3 = -1 \). The cross terms cancel as always, and the result is negative here, which is fine. Numerator: \( \sqrt{6}(\sqrt{2} - \sqrt{3}) = \sqrt{12} - \sqrt{18} \). Simplify each: \( \sqrt{12} = 2\sqrt{3} \) and \( \sqrt{18} = 3\sqrt{2} \), so the numerator is \( 2\sqrt{3} - 3\sqrt{2} \). The fraction is \( \dfrac{2\sqrt{3} - 3\sqrt{2}}{-1} = 3\sqrt{2} - 2\sqrt{3} \), distributing the negative to both terms. Check numerically: \( \sqrt{6} \approx 2.449 \) and \( \sqrt{2} + \sqrt{3} \approx 3.146 \), so the original is about 0.778. The answer gives \( 3(1.414) - 2(1.732) \approx 4.243 - 3.464 \approx 0.778 \). Correct. \( 3\sqrt{2} - 2\sqrt{3} \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \sqrt{x} = 6 \).
   Show the full solution

   \( x = 36 \)

2. Solve \( \sqrt{x - 2} = 5 \).
   Show the full solution

   \( x - 2 = 25 \). Check: \( \sqrt{25} = 5 \). \( x = 27 \)

3. Solve \( \sqrt{3x} = 9 \).
   Show the full solution

   \( 3x = 81 \). \( x = 27 \)

4. Does \( \sqrt{x} = -4 \) have a solution?
   Show the full solution

   No; a principal root is never negative

5. Why must every solution be checked?
   Show the full solution

   Squaring can create extraneous solutions

6. Solve \( \sqrt{x + 4} + 3 = 7 \).
   Show the full solution

   Isolate the radical first: \( \sqrt{x + 4} = 4 \). Square both sides: \( x + 4 = 16 \), so \( x = 12 \). Check: \( \sqrt{12 + 4} + 3 = \sqrt{16} + 3 = 4 + 3 = 7 \). Correct. Squaring before isolating would have given \( x + 4 + 6\sqrt{x+4} + 9 = 49 \), which still contains a radical and is worse than the original. \( x = 12 \)

7. Solve \( \sqrt{3x + 4} = x \).
   Show the full solution

   Square both sides: \( 3x + 4 = x^2 \), so \( x^2 - 3x - 4 = 0 \). Factoring: \( (x - 4)(x + 1) = 0 \), giving \( x = 4 \) or \( x = -1 \). Check \( x = 4 \): \( \sqrt{16} = 4 \), and the right side is 4. Correct. Check \( x = -1 \): \( \sqrt{1} = 1 \), but the right side is \( -1 \). Fails, so it is extraneous. \( x = 4 \) only

8. Solve \( \sqrt{x + 11} = x - 1 \).
   Show the full solution

   Square both sides, expanding the binomial fully: \( x + 11 = x^2 - 2x + 1 \), so \( 0 = x^2 - 3x - 10 \). Factoring: \( (x - 5)(x + 2) = 0 \), giving \( x = 5 \) or \( x = -2 \). Check \( x = 5 \): \( \sqrt{16} = 4 \), and \( 5 - 1 = 4 \). Correct. Check \( x = -2 \): \( \sqrt{9} = 3 \), but \( -2 - 1 = -3 \). Fails. Note the shortcut: the right side must be nonnegative, so \( x \geq 1 \), which rules out \( -2 \) in advance. \( x = 5 \) only

9. Explain exactly how squaring creates an extraneous solution, using \( \sqrt{x} = x - 2 \).
   Show the full solution

   Squaring gives \( x = x^2 - 4x + 4 \), so \( x^2 - 5x + 4 = 0 \), factoring as \( (x - 1)(x - 4) = 0 \) with candidates \( x = 1 \) and \( x = 4 \). Check \( x = 4 \): \( \sqrt{4} = 2 \) and \( 4 - 2 = 2 \). True, so it is a genuine solution. Check \( x = 1 \): \( \sqrt{1} = 1 \) but \( 1 - 2 = -1 \). The original equation claims \( 1 = -1 \), which is false. The mechanism is visible in that last line. Squaring both sides of \( 1 = -1 \) gives \( 1 = 1 \), a true statement. Squaring is not a reversible operation on equations: it maps both \( a = b \) and \( a = -b \) to the same squared equation, so the squared version has every solution the original had plus the solutions of the sign-flipped version. This is why the check must use the original equation. Checking in the squared equation would accept the extraneous answer, since it genuinely solves that one. \( x = 4 \) only; squaring turns the false statement \( 1 = -1 \) into the true statement \( 1 = 1 \)

10. Solve \( \sqrt{2x + 5} - \sqrt{x + 2} = 1 \).
    Show the full solution

    Isolate one radical: \( \sqrt{2x + 5} = 1 + \sqrt{x + 2} \). Square both sides, treating the right side as a binomial: \( 2x + 5 = 1 + 2\sqrt{x + 2} + (x + 2) = x + 3 + 2\sqrt{x+2} \). One radical remains, so isolate it again: \( 2x + 5 - x - 3 = 2\sqrt{x + 2} \), giving \( x + 2 = 2\sqrt{x + 2} \). Square again: \( (x + 2)^2 = 4(x + 2) \), so \( x^2 + 4x + 4 = 4x + 8 \), giving \( x^2 = 4 \) and \( x = \pm 2 \). Check \( x = 2 \): \( \sqrt{9} - \sqrt{4} = 3 - 2 = 1 \). Correct. Check \( x = -2 \): \( \sqrt{1} - \sqrt{0} = 1 - 0 = 1 \). Also correct, so both survive. Two radicals require two rounds of squaring, with an isolation step before each. And the check here kept both answers, which is a reminder that extraneous solutions are a possibility to test for, not an automatic outcome. \( x = 2 \) and \( x = -2 \), both genuine

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \dfrac{6x^2}{2x} \).
   Show the full solution

   \( 3x \), for \( x \neq 0 \)

2. What value is excluded from \( \dfrac{5}{x - 4} \)?
   Show the full solution

   \( x = 4 \)

3. Simplify \( \dfrac{x^2 - 4}{x + 2} \).
   Show the full solution

   \( \dfrac{(x+2)(x-2)}{x+2} \). \( x - 2 \), for \( x \neq -2 \)

4. Can you cancel the \( x \) in \( \dfrac{x + 4}{x} \)?
   Show the full solution

   No; the \( x \) in the numerator is a term, not a factor

5. What values are excluded from \( \dfrac{3}{x^2 - 9} \)?
   Show the full solution

   \( (x+3)(x-3) = 0 \). \( x = 3 \) and \( x = -3 \)

6. Simplify \( \dfrac{x^2 + 5x + 6}{x^2 - 4} \) with exclusions.
   Show the full solution

   Numerator: two numbers multiplying to 6 and adding to 5 are 2 and 3, so \( (x + 2)(x + 3) \). Denominator: a difference of squares, \( (x + 2)(x - 2) \). Exclusions from the original denominator: \( x \neq -2 \) and \( x \neq 2 \). Cancel \( (x + 2) \): the result is \( \dfrac{x + 3}{x - 2} \). Check at \( x = 0 \): the original is \( \dfrac{6}{-4} = -1.5 \), and the simplified form is \( \dfrac{3}{-2} = -1.5 \). Correct. \( \dfrac{x + 3}{x - 2} \), for \( x \neq \pm 2 \)

7. Simplify \( \dfrac{2x^2 - 8}{x^2 + 4x + 4} \).
   Show the full solution

   Numerator: factor the GCF first, \( 2(x^2 - 4) = 2(x + 2)(x - 2) \). Denominator: a perfect square trinomial, \( (x + 2)^2 \). Exclusion: \( x \neq -2 \). Cancel one \( (x + 2) \): \( \dfrac{2(x - 2)}{x + 2} \). Check at \( x = 0 \): the original is \( \dfrac{-8}{4} = -2 \), and the simplified form is \( \dfrac{2(-2)}{2} = -2 \). Correct. \( \dfrac{2(x - 2)}{x + 2} \), for \( x \neq -2 \)

8. Simplify \( \dfrac{x - 5}{5 - x} \).
   Show the full solution

   The two parts look unrelated until one is rewritten. Factor \( -1 \) out of the denominator: \( 5 - x = -(x - 5) \). So the expression is \( \dfrac{x - 5}{-(x - 5)} = -1 \), for \( x \neq 5 \). Check at \( x = 7 \): \( \dfrac{2}{-2} = -1 \). Correct. At \( x = 0 \): \( \dfrac{-5}{5} = -1 \). Correct. Any pair of expressions that are opposites of each other divides to \( -1 \), and spotting that saves considerable work. \( -1 \), for \( x \neq 5 \)

9. A student simplifies \( \dfrac{x^2 + 9}{x + 3} \) to \( x + 3 \). Explain the error.
   Show the full solution

   They treated \( x^2 + 9 \) as though it factored into \( (x+3)(x+3) \), but a sum of squares does not factor over the integers, as lesson 7.7 established. Expanding \( (x+3)^2 \) gives \( x^2 + 6x + 9 \), which has a middle term the numerator does not have. A numerical check exposes it: at \( x = 1 \) the original is \( \dfrac{10}{4} = 2.5 \), while their answer gives 4. The expression \( \dfrac{x^2 + 9}{x + 3} \) is already in simplest form, since the numerator has no factor in common with the denominator. Its only exclusion is \( x \neq -3 \). Not every rational expression simplifies, and reporting that one does not is the correct answer when it is true. It does not simplify; \( x^2 + 9 \) is prime, and the correct answer is the original with \( x \neq -3 \)

10. Simplify \( \dfrac{x^3 - 4x}{x^2 - x - 6} \) with exclusions, and explain why an exclusion is kept after its factor cancels.
    Show the full solution

    Numerator: factor the GCF, \( x(x^2 - 4) = x(x + 2)(x - 2) \). Denominator: two numbers multiplying to \( -6 \) and adding to \( -1 \) are \( -3 \) and 2, so \( (x - 3)(x + 2) \). Exclusions from the original denominator: \( x \neq 3 \) and \( x \neq -2 \). Cancel \( (x + 2) \): the result is \( \dfrac{x(x - 2)}{x - 3} \). Check at \( x = 1 \): the original is \( \dfrac{1 - 4}{1 - 1 - 6} = \dfrac{-3}{-6} = 0.5 \), and the simplified form is \( \dfrac{1(-1)}{-2} = 0.5 \). Correct. The exclusion \( x \neq -2 \) stays because the two expressions are equal only where both are defined. At \( x = -2 \) the original has a denominator of zero and no value at all, while the simplified form gives \( \dfrac{(-2)(-4)}{-5} = -1.6 \). Simplifying produced an expression that is defined at a point where the original was not, so the restriction has to be carried along to keep the two genuinely equal. On a graph this shows up as a hole at \( x = -2 \). \( \dfrac{x(x - 2)}{x - 3} \), for \( x \neq 3 \) and \( x \neq -2 \); the canceled factor still made the original undefined

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Simplify \( \dfrac{2}{x} \cdot \dfrac{x}{5} \).
   Show the full solution

   \( \frac{2}{5} \), for \( x \neq 0 \)

2. Simplify \( \dfrac{3}{x} \div \dfrac{6}{x^2} \).
   Show the full solution

   \( \dfrac{3}{x} \cdot \dfrac{x^2}{6} = \dfrac{3x}{6} \). \( \frac{x}{2} \), for \( x \neq 0 \)

3. Simplify \( \dfrac{2}{x} + \dfrac{3}{x} \).
   Show the full solution

   The denominators already match. \( \frac{5}{x} \)

4. What is the least common denominator of \( \dfrac{1}{x} \) and \( \dfrac{1}{x + 1} \)?
   Show the full solution

   \( x(x + 1) \)

5. How do you divide by a rational expression?
   Show the full solution

   Multiply by its reciprocal

6. Simplify \( \dfrac{x^2 - 1}{x + 4} \cdot \dfrac{x + 4}{x - 1} \).
   Show the full solution

   Factor: \( x^2 - 1 = (x + 1)(x - 1) \). \( \dfrac{(x+1)(x-1)}{x+4} \cdot \dfrac{x+4}{x-1} \). Exclusions: \( x \neq -4 \) and \( x \neq 1 \). Cancel \( (x + 4) \) and \( (x - 1) \), leaving \( x + 1 \). Check at \( x = 0 \): the original is \( \dfrac{-1}{4} \cdot \dfrac{4}{-1} = 1 \), and the answer is \( 0 + 1 = 1 \). Correct. \( x + 1 \), for \( x \neq -4 \) and \( x \neq 1 \)

7. Simplify \( \dfrac{4}{x - 3} - \dfrac{2}{x} \).
   Show the full solution

   The denominators share no factor, so the least common denominator is \( x(x - 3) \). Exclusions: \( x \neq 3 \) and \( x \neq 0 \). Rewrite: \( \dfrac{4x}{x(x-3)} - \dfrac{2(x - 3)}{x(x-3)} \). Combine, distributing the minus to both terms: \( \dfrac{4x - 2x + 6}{x(x-3)} = \dfrac{2x + 6}{x(x - 3)} \). Check at \( x = 1 \): the original is \( \dfrac{4}{-2} - 2 = -4 \), and the answer is \( \dfrac{8}{-2} = -4 \). Correct. \( \dfrac{2x + 6}{x(x - 3)} \)

8. Simplify \( \dfrac{x^2 - 9}{x^2} \div \dfrac{x + 3}{x} \).
   Show the full solution

   Multiply by the reciprocal: \( \dfrac{x^2 - 9}{x^2} \cdot \dfrac{x}{x + 3} \). Factor: \( \dfrac{(x+3)(x-3)}{x^2} \cdot \dfrac{x}{x+3} \). Exclusions: \( x \neq 0 \) and \( x \neq -3 \). Cancel \( (x + 3) \), and cancel one \( x \) from \( x^2 \) against the \( x \) on top: \( \dfrac{x - 3}{x} \). Check at \( x = 1 \): the original is \( \dfrac{-8}{1} \div \dfrac{4}{1} = -2 \), and the answer is \( \dfrac{-2}{1} = -2 \). Correct. \( \dfrac{x - 3}{x} \)

9. A student writes \( \dfrac{2}{x} + \dfrac{3}{y} = \dfrac{5}{x + y} \). Explain the error.
   Show the full solution

   They added numerators and denominators separately, which is not how fraction addition works. Testing with numbers settles it: at \( x = 2 \) and \( y = 3 \), the left side is \( 1 + 1 = 2 \), while their right side is \( \dfrac{5}{5} = 1 \). The correct procedure uses a common denominator of \( xy \): \( \dfrac{2y}{xy} + \dfrac{3x}{xy} = \dfrac{2y + 3x}{xy} \). Checking at \( x = 2 \), \( y = 3 \): \( \dfrac{6 + 6}{6} = 2 \). Correct. The reason denominators cannot simply be added is that a denominator says what size the pieces are, not how many there are. Two different sizes must first be rewritten as a common size before the counts can be combined. \( \dfrac{2y + 3x}{xy} \); denominators are never added

10. Simplify \( \dfrac{x}{x^2 - x - 6} + \dfrac{2}{x - 3} \) with exclusions.
    Show the full solution

    Factor the first denominator: two numbers multiplying to \( -6 \) and adding to \( -1 \) are \( -3 \) and 2, so \( x^2 - x - 6 = (x - 3)(x + 2) \). The denominators are \( (x - 3)(x + 2) \) and \( (x - 3) \). The second is already a factor of the first, so the least common denominator is \( (x - 3)(x + 2) \). Exclusions: \( x \neq 3 \) and \( x \neq -2 \). Rewrite the second fraction by multiplying top and bottom by \( (x + 2) \): \( \dfrac{x}{(x-3)(x+2)} + \dfrac{2(x + 2)}{(x-3)(x+2)} \). Combine the numerators: \( x + 2x + 4 = 3x + 4 \). The result is \( \dfrac{3x + 4}{(x - 3)(x + 2)} \). The numerator does not factor in a way that matches either denominator factor, so nothing cancels. Check at \( x = 0 \): the original is \( \dfrac{0}{-6} + \dfrac{2}{-3} \approx -0.667 \), and the answer is \( \dfrac{4}{-6} \approx -0.667 \). Correct. \( \dfrac{3x + 4}{(x - 3)(x + 2)} \), for \( x \neq 3 \) and \( x \neq -2 \)

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Solve \( \dfrac{x}{3} = 4 \).
   Show the full solution

   \( x = 12 \)

2. Solve \( \dfrac{6}{x} = 2 \).
   Show the full solution

   \( 6 = 2x \). \( x = 3 \)

3. Solve \( \dfrac{2}{x} = \dfrac{8}{12} \).
   Show the full solution

   Cross multiply: \( 24 = 8x \). \( x = 3 \)

4. What value is excluded from \( \dfrac{5}{x - 2} = 1 \)?
   Show the full solution

   \( x = 2 \)

5. Solve that equation.
   Show the full solution

   \( 5 = x - 2 \), and 7 is not excluded. \( x = 7 \)

6. Solve \( \dfrac{x}{4} + \dfrac{1}{2} = \dfrac{5}{4} \).
   Show the full solution

   No variable appears in a denominator, so there are no exclusions. The least common denominator is 4. Multiply every term by 4: \( x + 2 = 5 \), so \( x = 3 \). Check: \( \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{2}{4} = \dfrac{5}{4} \). Correct. \( x = 3 \)

7. Solve \( \dfrac{4}{x} + \dfrac{1}{3} = \dfrac{5}{3} \).
   Show the full solution

   Exclusion: \( x \neq 0 \). The least common denominator is \( 3x \). Multiply every term: \( 12 + x = 5x \), so \( 12 = 4x \) and \( x = 3 \). Not excluded, so it stands. Check: \( \dfrac{4}{3} + \dfrac{1}{3} = \dfrac{5}{3} \). Correct. \( x = 3 \)

8. Solve \( \dfrac{2}{x - 1} = \dfrac{3}{x + 2} \).
   Show the full solution

   Exclusions: \( x \neq 1 \) and \( x \neq -2 \). Cross multiply: \( 2(x + 2) = 3(x - 1) \), so \( 2x + 4 = 3x - 3 \), giving \( 7 = x \). Not excluded, so it stands. Check: \( \dfrac{2}{6} = \dfrac{1}{3} \) and \( \dfrac{3}{9} = \dfrac{1}{3} \). Correct. \( x = 7 \)

9. Solve \( \dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 3 \) and explain the result.
   Show the full solution

   Exclusion: \( x \neq 2 \). The least common denominator is \( x - 2 \). Multiply every term, including the 3: \( x = 2 + 3(x - 2) \), so \( x = 2 + 3x - 6 = 3x - 4 \), giving \( 4 = 2x \) and \( x = 2 \). The only candidate is the excluded value. Substituting into the original gives \( \dfrac{2}{0} \), which is undefined, so it cannot be a solution. The equation therefore has no solution. What has happened is that clearing denominators produced an equation that is not fully equivalent to the original: it permits \( x = 2 \), which the original never did, because multiplying by \( x - 2 \) is multiplying by zero when \( x = 2 \). Listing the exclusion first made the outcome clear instead of puzzling. No solution; the only candidate is the excluded value \( x = 2 \)

10. A crew paints a room in 6 hours alone, and with a helper in 4 hours. How long would the helper take alone?
    Show the full solution

    Work problems are set up in rates: what fraction of the job each does per hour. The crew does \( \dfrac{1}{6} \) of the room per hour. Let \( h \) be the hours the helper would need alone, so the helper does \( \dfrac{1}{h} \) per hour. Working together they do \( \dfrac{1}{4} \) per hour, since they finish in 4 hours. Rates add: \( \dfrac{1}{6} + \dfrac{1}{h} = \dfrac{1}{4} \). Exclusion: \( h \neq 0 \), which is also physically obvious. The least common denominator of 6, \( h \) and 4 is \( 12h \). Multiplying every term: \( 2h + 12 = 3h \), so \( h = 12 \). Check: \( \dfrac{1}{6} + \dfrac{1}{12} = \dfrac{2}{12} + \dfrac{1}{12} = \dfrac{3}{12} = \dfrac{1}{4} \). Correct. Check against the situation as well: the helper is slower than the crew, taking 12 hours against 6, which is consistent with the pair together taking 4 hours rather than the 3 they would take if equally fast. 12 hours

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Find the mean of 3, 5, 10.
   Show the full solution

   \( 18 \div 3 \). 6

2. Find the median of 8, 3, 5.
   Show the full solution

   Ordered: 3, 5, 8. 5

3. Find the mode of 2, 4, 4, 7.
   Show the full solution

   4

4. Find the range of 11, 4, 19, 7.
   Show the full solution

   \( 19 - 4 \). 15

5. What does a standard deviation of 0 mean?
   Show the full solution

   Every value is identical

6. Find the mean and median of 6, 9, 11, 14, 20.
   Show the full solution

   The data is ordered and has 5 values. Sum: \( 6 + 9 + 11 + 14 + 20 = 60 \), so the mean is \( 60 \div 5 = 12 \). With an odd count the median is the middle value, the 3rd, which is 11. The mean exceeds the median slightly, reflecting the value 20 sitting further above the middle than 6 sits below it. Mean 12, median 11

7. Find the interquartile range of 3, 5, 8, 10, 12, 15, 18, 22.
   Show the full solution

   The data is ordered with 8 values, so split into halves of 4. Lower half: 3, 5, 8, 10. Its median is \( \dfrac{5 + 8}{2} = 6.5 \), so \( Q_1 = 6.5 \). Upper half: 12, 15, 18, 22. Its median is \( \dfrac{15 + 18}{2} = 16.5 \), so \( Q_3 = 16.5 \). \( \text{IQR} = 16.5 - 6.5 = 10 \). The middle half of the data spans 10 units, while the full range spans \( 22 - 3 = 19 \). 10

8. Find the standard deviation of 2, 4, 6, 8, 10.
   Show the full solution

   Mean: \( 30 \div 5 = 6 \). Deviations: \( -4, -2, 0, 2, 4 \). They sum to zero, as a check. Squares: 16, 4, 0, 4, 16, totaling 40. Average of the squares: \( 40 \div 5 = 8 \). Standard deviation: \( \sqrt{8} \approx 2.83 \). So a typical value lies about 2.83 units from the mean of 6, which is reasonable for data running from 2 to 10. About 2.83

9. Two classes both average 75. One has a standard deviation of 3 and the other 18. Describe the difference.
   Show the full solution

   The center is identical, so the mean alone cannot tell them apart. The spread is completely different. In the first class, a typical score sits about 3 points from 75, so nearly everyone scored somewhere in the high 60s to high 70s. The class performed uniformly, and a single number describes it well. In the second class, a typical score sits about 18 points from 75, so scores likely range from the 40s to near 100. That class contains both students who understood the material thoroughly and students who did not, and reporting only the average of 75 conceals a genuine problem. The practical consequence is that the two classes need different responses. The first might move on together; the second needs its two groups addressed separately. This is why spread is reported alongside center rather than as an optional extra. Same center, very different consistency: the first class is tightly grouped, the second contains widely differing performances

10. For 5, 5, 5, 5, 25, compute the mean, median and standard deviation, and say which summary is most misleading.
    Show the full solution

    Mean: \( 45 \div 5 = 9 \). Median: with 5 ordered values the middle is the 3rd, which is 5. Standard deviation: deviations from 9 are \( -4, -4, -4, -4, 16 \), summing to zero as required. Squares: 16, 16, 16, 16, 256, totaling 320. Average: \( 320 \div 5 = 64 \). Standard deviation: \( \sqrt{64} = 8 \). The mean of 9 is the most misleading summary. No value in the data set is anywhere near 9: four of the five are exactly 5, and the other is 25. The mean describes a typical value that does not exist, because the single large value pulled it away from the bulk of the data. The median of 5 describes the data far better, since it is the value four of the five points actually take. And the standard deviation of 8 is doing useful work by warning that the spread is almost as large as the mean itself, which signals that the mean should not be trusted on its own. Lesson 11.2 takes up this situation directly. Mean 9, median 5, standard deviation 8; the mean is most misleading because a single extreme value pulls it away from the bulk of the data

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

1. Which measure of center is resistant to outliers?
   Show the full solution

   The median

2. If data is skewed right, which is larger, the mean or the median?
   Show the full solution

   The mean

3. What multiplier does the interquartile range outlier rule use?
   Show the full solution

   1.5

4. If \( Q_1 = 10 \) and \( Q_3 = 20 \), what is the upper fence?
   Show the full solution

   \( \text{IQR} = 10 \), so \( 20 + 15 \). 35

5. Which summary suits skewed data?
   Show the full solution

   The median with the interquartile range

6. For 4, 6, 7, 8, 9, 11, 40, test whether 40 is an outlier.
   Show the full solution

   Seven ordered values, so the median is the 4th, which is 8. Lower half: 4, 6, 7, so \( Q_1 = 6 \). Upper half: 9, 11, 40, so \( Q_3 = 11 \). \( \text{IQR} = 11 - 6 = 5 \), and \( 1.5 \times 5 = 7.5 \). Upper fence: \( 11 + 7.5 = 18.5 \). Lower fence: \( 6 - 7.5 = -1.5 \). Since \( 40 \gt 18.5 \), it is an outlier. No value falls below \( -1.5 \), so there is no low outlier. Yes, 40 exceeds the upper fence of 18.5

7. Compute the mean of that data set with and without 40.
   Show the full solution

   With 40: the sum is \( 4 + 6 + 7 + 8 + 9 + 11 + 40 = 85 \), so the mean is \( 85 \div 7 \approx 12.14 \). Without 40: the sum is 45 over 6 values, so the mean is \( 45 \div 6 = 7.5 \). The mean dropped by about 4.6, which is a large shift for removing one value from seven. Meanwhile the median moves only from 8 to \( \dfrac{7+8}{2} = 7.5 \), a change of 0.5. About 12.14 with it, 7.5 without it

8. House prices in a town have a mean of $420,000 and a median of $310,000. Describe the distribution.
   Show the full solution

   The mean sits well above the median, by $110,000, which is the signature of a distribution skewed to the right. The interpretation is that most houses sell for prices near the median, around $310,000, but a smaller number of very expensive houses form a long tail of high values. Those high prices enter the mean at their full size and pull it upward, while the median only counts them as being above the middle. A buyer asking what a typical house costs should be told $310,000. Quoting the mean of $420,000 would overstate what most of the market looks like, even though the figure itself is correct. Skewed right: most houses are near $310,000, with a few expensive homes pulling the mean up

9. Explain why the median barely moves when an outlier is made more extreme.
   Show the full solution

   The median is determined by position in the ordered list, not by magnitude. Once a value is known to be the largest, making it larger still does not change its position, so the value sitting in the middle is unaffected. Take 2, 5, 9, 11, 40. The median is the 3rd value, 9. Change 40 to 400, or to 4,000,000, and the median is still 9, because the middle position still holds the same number. The mean behaves completely differently, because it divides a total that includes the outlier at full size. For the same three versions the means are 13.4, 85.4 and about 800,005.4. One value has moved the mean arbitrarily far. This is precisely what makes the median the safer summary when a data set may contain recording errors or genuine extremes, and it is the reason both are taught rather than just the mean. The median depends on position, and an outlier keeps the same position however extreme it becomes

10. A data set of test scores contains a zero from a student who was absent. Discuss whether to include it.
    Show the full solution

    The answer depends on what question the summary is supposed to answer, and the choice must be stated rather than made silently. If the question is how well students understood the material, the zero should be excluded, because it measures attendance rather than understanding. Including it makes the class look less capable than it is, and the exclusion should be reported: "one absent student excluded." If the question is what fraction of possible points the class actually earned, perhaps for a grade book total, the zero belongs, because the points genuinely were not earned. What is not acceptable is removing the zero without saying so. A reader given a mean has no way to know that a value was dropped, and undisclosed deletion of inconvenient data is the difference between analysis and misrepresentation. A reasonable practice is to report both: the mean with the zero, the mean without it, and a sentence explaining the difference. That lets the reader judge, which is the purpose of a summary in the first place. It depends on the question being answered, and whichever choice is made must be disclosed