Topic 1.3 · Unit 1 · CA CCSS A-REI.1, A-REI.3, A-CED.4

Solving linear equations and rearranging formulas

Solving for $ x $ and rearranging a formula for one of its letters are the same operation, undo what is being done to the target, in reverse order, doing it to both sides. Treating them as two separate skills is why the formula questions feel harder than they are.

The method

Clear fractions first by multiplying every term by the common denominator. Every term, including ones without a fraction.
Distribute across any brackets.
Collect the target on one side and everything else on the other. Move the smaller coefficient to avoid negatives if you can choose.
Factor out the target if it appears in more than one term; this is the step formulas need and simple equations do not. From $ ax + bx = c $ you get $ x(a+b) = c $, so $ x = \frac{c}{a+b} $.
Divide by the whole coefficient and check by substituting back.

Special cases worth recognizing instantly: if the variables cancel and you are left with something true, such as $ 6 = 6 $, every number works, infinitely many solutions. If you are left with something false, such as $ 6 = 9 $, no number works, no solution. Students often write "0" for the second; the answer is that there is no solution, which is not the same as $ x = 0 $.

Worked examples

Example 1: variables on both sides with fractions. Solve $ \frac{x}{3} + 2 = \frac{x}{2} - 1 $.

Common denominator 6, multiply every term: $ 2x + 12 = 3x - 6 $. Then $ 12 + 6 = 3x - 2x $, so $ x = 18 $.

Check: $ \frac{18}{3} + 2 = 8 $ and $ \frac{18}{2} - 1 = 8 $. ✓

Example 2: no solution against infinitely many. Solve $ 4(x - 2) = 4x + 5 $.

$ 4x - 8 = 4x + 5 $. Subtracting $ 4x $ from both sides gives $ -8 = 5 $, which is false. The variable has vanished and left a false statement, so no solution. Had it left $ -8 = -8 $, the answer would be all real numbers.

Example 3: a formula where the target appears twice. Make $ x $ the subject of $ ax + b = cx + d $.

Collect the $ x $ terms: $ ax - cx = d - b $. Now factor, which is the step that makes this solvable: $ x(a - c) = d - b $, so $ x = \frac{d - b}{a - c} $. Without factoring there is no way to isolate a letter that appears in two terms.

Practice · 15 problems

1–6 are equations, 7–11 mix in fractions and special cases, 12–15 are formula rearrangements and word problems.

Solve $ 5x - 7 = 23 $.
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$ 5x = 30 $, so $ x = 6 $. $ x = 6 $
Solve $ 3x + 4 = x + 16 $.
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$ 3x - x = 16 - 4 $, so $ 2x = 12 $ and $ x = 6 $. $ x = 6 $
Solve $ 2(x + 5) = 18 $.
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Distribute: $ 2x + 10 = 18 $, so $ 2x = 8 $ and $ x = 4 $. Dividing both sides by 2 first is quicker and equally valid. $ x = 4 $
Solve $ 7 - 3x = 22 $.
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$ -3x = 15 $, so $ x = -5 $. Dividing by a negative flips the sign of the answer. $ x = -5 $
Solve $ 4(2x - 1) = 3(x + 7) $.
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$ 8x - 4 = 3x + 21 $, so $ 5x = 25 $ and $ x = 5 $. $ x = 5 $
Solve $ \frac{x}{4} = 9 $.
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Multiply both sides by 4: $ x = 36 $. $ x = 36 $
Solve $ \frac{2x}{3} - 1 = \frac{x}{2} + 2 $.
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Multiply every term by 6: $ 4x - 6 = 3x + 12 $, so $ x = 18 $. Forgetting to multiply the $ -1 $ and $ +2 $ is the usual slip. $ x = 18 $
Solve $ 3(x - 2) = 3x - 6 $.
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$ 3x - 6 = 3x - 6 $. The variable cancels leaving a true statement, so every real number satisfies it. Infinitely many solutions
Solve $ 2x + 5 = 2x - 3 $.
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Subtracting $ 2x $ gives $ 5 = -3 $, which is false. No solution: not $ x = 0 $.
Solve $ 0.5x + 1.2 = 3.7 $.
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$ 0.5x = 2.5 $, so $ x = 5 $. Multiplying through by 10 first to clear decimals works equally well. $ x = 5 $
Solve $ \frac{x + 3}{5} = 4 $.
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Multiply both sides by 5: $ x + 3 = 20 $, so $ x = 17 $. The whole numerator is being divided, so the whole numerator gets multiplied. $ x = 17 $
Make $ h $ the subject of $ A = \frac{1}{2}bh $.
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Multiply by 2: $ 2A = bh $. Divide by $ b $: $ h = \frac{2A}{b} $. $ h = \frac{2A}{b} $
Make $ C $ the subject of $ F = \frac{9}{5}C + 32 $.
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$ F - 32 = \frac{9}{5}C $, then multiply by $ \frac{5}{9} $: $ C = \frac{5}{9}(F - 32) $. $ C = \frac{5}{9}(F - 32) $
Make $ x $ the subject of $ mx + n = px - q $.
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Collect: $ mx - px = -q - n $. Factor, the step this type needs: $ x(m - p) = -(q + n) $, so $ x = \frac{-(q+n)}{m-p} $, which can also be written $ \frac{q+n}{p-m} $. $ x = \frac{q+n}{p-m} $
A plumber charges a $60 call-out plus $45 per hour. A job cost $217.50. How many hours did it take?
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Set up $ 60 + 45h = 217.50 $. Then $ 45h = 157.50 $ and $ h = 3.5 $. 3.5 hours

Solving linear equations and rearranging formulas

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