Topic 1.4 · Unit 1 · CA CCSS A-REI.3
Linear inequalities in one variable
Solve an inequality exactly as you would the matching equation, with one extra rule that accounts for most of the lost marks on this topic: multiplying or dividing both sides by a negative reverses the inequality sign.
- Solve as an equation: clear fractions, distribute, collect, divide.
- Flip the sign if and only if you multiply or divide both sides by a negative number. Adding or subtracting a negative does not flip it.
- Graph it. Open circle for \( < \) or \( > \), closed circle for \( \le \) or \( \ge \). Shade toward the numbers that work.
- Test one value from the shaded region in the original inequality. This catches a flipped sign immediately and takes ten seconds.
- Compound inequalities: "and" means the overlap of the two regions; "or" means both regions together. \( -3 < x \le 5 \) is an "and" written compactly.
Where marks are lost: \( -2x > 8 \) gives \( x < -4 \), not \( x > -4 \). Test it: \( x = -5 \) gives \( -2(-5) = 10 > 8 \) ✓, while \( x = 0 \) gives \( 0 > 8 \) ✗. Avoid the rule entirely by moving the variable to whichever side keeps it positive.
Example 1: the flip. Solve \( 5 - 3x \ge 17 \).
\( -3x \ge 12 \). Dividing by \( -3 \) reverses the sign: \( x \le -4 \).
Check with \( x = -5 \): \( 5 - 3(-5) = 20 \ge 17 \) ✓. Check with \( x = 0 \): \( 5 \ge 17 \) ✗. The solution set is correct.
Example 2: avoiding the flip. Solve \( 4 - 2x < 10 \) by moving the variable instead.
Add \( 2x \) to both sides: \( 4 < 10 + 2x \). Subtract 10: \( -6 < 2x \). Divide by positive 2: \( -3 < x \), that is \( x > -3 \). No flip was ever needed, because the variable was never divided by a negative.
Example 3: a compound inequality. Solve \( -7 < 2x + 1 \le 9 \).
Operate on all three parts at once. Subtract 1: \( -8 < 2x \le 8 \). Divide by 2: \( -4 < x \le 4 \). Graph with an open circle at \( -4 \), a closed circle at 4, and shading between.
1–6 are direct, 7–11 involve the flip or fractions, and 12–15 are compound inequalities and word problems.
- Solve \( x + 7 < 12 \).
Show the full solution
Subtract 7. No multiplication by a negative, so no flip. \( x < 5 \)
- Solve \( 3x \ge 21 \).
Show the full solution
Divide by positive 3. \( x \ge 7 \)
- Solve \( 2x - 5 > 9 \).
Show the full solution
\( 2x > 14 \), so \( x > 7 \). \( x > 7 \)
- Solve \( -4x < 20 \).
Show the full solution
Dividing by \( -4 \) flips the sign: \( x > -5 \). Test \( x = 0 \): \( 0 < 20 \) ✓. \( x > -5 \)
- Solve \( 6 - x \le 2 \).
Show the full solution
\( -x \le -4 \); dividing by \( -1 \) flips: \( x \ge 4 \). Or move the \( x \) across instead: \( 6 \le 2 + x \), giving \( 4 \le x \). \( x \ge 4 \)
- Graph the solution to \( x \ge -2 \) on a number line.
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A closed circle at \( -2 \), because \( \ge \) includes the endpoint, with shading to the right. Closed circle at −2, shaded right
- Solve \( 5 - 2x > 13 \).
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\( -2x > 8 \), dividing by \( -2 \) flips: \( x < -4 \). Test \( x = -5 \): \( 5 + 10 = 15 > 13 \) ✓. \( x < -4 \)
- Solve \( 3(x - 2) \le 2x + 1 \).
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\( 3x - 6 \le 2x + 1 \), so \( x \le 7 \). No flip, nothing was divided by a negative. \( x \le 7 \)
- Solve \( \frac{x}{3} + 2 > 5 \).
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\( \frac{x}{3} > 3 \), multiply by positive 3: \( x > 9 \). \( x > 9 \)
- Solve \( \frac{2 - x}{4} \ge 1 \).
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Multiply by 4: \( 2 - x \ge 4 \), so \( -x \ge 2 \) and dividing by \( -1 \) flips to \( x \le -2 \). \( x \le -2 \)
- A student solves \( -3x < 9 \) and writes \( x < -3 \). Find the error.
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They divided by \( -3 \) without reversing the sign. It should be \( x > -3 \). Testing \( x = 0 \) exposes it at once: \( 0 < 9 \) is true, so 0 must be in the solution set, and \( x < -3 \) excludes it. \( x > -3 \)
- Solve \( -1 \le 3x + 2 < 11 \).
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Subtract 2 from all three parts: \( -3 \le 3x < 9 \). Divide all by 3: \( -1 \le x < 3 \). \( -1 \le x < 3 \)
- Solve \( x - 1 < 3 \) or \( x + 2 > 10 \).
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Separately: \( x < 4 \) or \( x > 8 \). "Or" keeps both regions, and they do not overlap, so the solution is two separate pieces. \( x < 4 \) or \( x > 8 \)
- A lift holds at most 600 kg. Four people weighing a total of 290 kg are inside. Boxes
weigh 25 kg each. How many boxes can be loaded?
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\( 290 + 25b \le 600 \), so \( 25b \le 310 \) and \( b \le 12.4 \). Boxes are whole, so at most 12. Rounding 12.4 up to 13 would exceed the limit, context decides the rounding direction. 12 boxes
- A student needs a mean of at least 80 across four tests. They have scored 76, 84 and 79.
What must they score on the fourth?
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\( \frac{76 + 84 + 79 + x}{4} \ge 80 \), so \( 239 + x \ge 320 \) and \( x \ge 81 \). At least 81