Math 8 · 8.EE.C.7B · A.REI.B.3

Multi-step equations

An equation has brackets, several terms with the unknown in them, or fractions and decimals everywhere. How do you turn it into one you already know how to solve?

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Two-step equationsMath 7 Greatest common factor and least common multipleGrades 5 to 6 Multiplying decimalsGrade 5 The distributive property with variablesMath 7 +1 more

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Absolute value equationsAlgebra 1 Equations with variables on both sidesMath 8 Solving radical equationsAlgebra 2

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When you reach for it

A two-step equation such as 3x + 7 = 25 has the unknown in a single term, and two moves undo what was done to it. Many equations arrive less tidy than that. In 5x + 3 - 2x + 4 = 25 the unknown appears in two terms. In 4(y - 3) + 2y = 30 part of it is inside a bracket. In \displaystyle \frac{x}{3} + \frac{x}{4} = 14 it sits under fractions.

Each of these still has the unknown on one side only, and each can be brought back to the two-step shape

px + q = r

by tidying the side the unknown is on. An equation that needs this tidying first is called a multi-step equation. Once it has the shape px + q = r, the two-step method does the rest. The new work is the tidying, and every tidying step must be one that cannot change the answer.

This lesson takes two days. Part 1 distributes brackets and combines like terms, then solves and checks. Part 2 clears fractions and decimals before solving, and builds such equations from stories: a bill that adds several charges, a perimeter that adds four sides.

The procedure

Part 1: tidy each side, then solve

Take 5x + 3 - 2x + 4 = 25. Its left side is a sum of four terms. A term is one of the pieces of a sum, joined to the others by + or -, and the sign in front of it belongs to it. A term may be a number such as 3, a product such as -2x, or a product with a bracket such as 4(y - 3). Like terms have the same letter part: 5x and -2x are like terms, and so are the constants 3 and 4.

Combine like terms. The commutative property of addition lets the terms be reordered, the associative property lets them be regrouped, and the distributive property gives 5x - 2x = (5 - 2)x = 3x. This is combining like terms:

5x + 3 - 2x + 4 = 3x + 7

So the equation is 3x + 7 = 25, a two-step equation. Subtract 7 from both sides to get 3x = 18, then divide both sides by 3 to get x = 6.

Distribute first when there is a bracket. In 4(y - 3) + 2y = 30, the y inside the bracket cannot be combined with 2y while the bracket is there. The distributive property multiplies the 4 into every term inside: 4(y - 3) = 4y - 12. Then the like terms can meet:

\begin{aligned} 4y - 12 + 2y &= 30 \\ 6y - 12 &= 30 \\ 6y &= 42 \\ y &= 7 \end{aligned}

The two moves together, distributing and then combining, are practiced in simplifying with the distributive property.

The sign in front of a bracket is part of the number that multiplies it. In 10 - 2(n - 6) the bracket is multiplied by -2, so -2(n - 6) = -2n + 12, since (-2)(-6) = 12. A bracket with only a minus sign in front is multiplied by -1: -(w + 4) = -w - 4.

Why tidying cannot change the answer. The properties of equality change both sides of an equation by the same amount. Tidying a side is a different kind of move: it does not change that side’s value at all. The expressions 4(y - 3) + 2y and 6y - 12 are equivalent: they give the same number for every value of y. So a number that makes 4(y - 3) + 2y = 30 true also makes 6y - 12 = 30 true, and the other way round. The two equations have exactly the same solutions.

Check in the original equation. Put y = 7 into 4(y - 3) + 2y = 30, not into the tidied 6y - 12 = 30: 4(4) + 14 = 16 + 14 = 30. A check in the tidied equation tests the solving but not the tidying, so a slip in distributing would pass unseen. Putting the number back where it started is what shows it is a solution.

When a bracket is the whole of one side, as in 3(x + 4) = 27, dividing first also works; that shape has its own lesson.

Part 2: clearing fractions and decimals

Fractions can be carried through every step, but it is often easier to remove them first. Take \displaystyle \frac{x}{3} + \frac{x}{4} = 14. The denominators are 3 and 4, and their least common multiple is 12. Multiply both sides by 12. The multiplication property of equality allows it, and since 12 is not 0 the step can be undone by dividing by 12, so the new equation has the same solutions as the old one. On the left, the distributive property carries the 12 to every term:

12 \cdot \frac{x}{3} + 12 \cdot \frac{x}{4} = 12 \cdot 14

Since 3 divides 12, \displaystyle 12 \cdot \frac{x}{3} = 4x, and likewise \displaystyle 12 \cdot \frac{x}{4} = 3x. Every fraction is gone:

\begin{aligned} 4x + 3x &= 168 \\ 7x &= 168 \\ x &= 24 \end{aligned}

Check: \displaystyle \frac{24}{3} + \frac{24}{4} = 8 + 6 = 14. Any common multiple of the denominators would clear them (24 would do); the least one keeps the numbers smallest.

Clearing decimals. A decimal is a fraction whose denominator is a power of ten: \displaystyle 0.35 = \frac{35}{100}. So multiplying both sides by 10, 100 or 1000, enough for the decimal with the most decimal places, turns every decimal into a whole number. The arithmetic is multiplying a decimal by 10 or 100, which moves its digits one or two places. In 1.2a - 0.35 + 0.8a = 2.65 the most places is two, so multiply by 100.

From words to an equation. Name the unknown with its unit (“let h be the number of hours”). Find the quantity the story tells you two ways, such as a total bill, a perimeter or a sum: once as an expression with several terms, once as a number. Set them equal, tidy, solve, and check the answer in the story itself, which catches an equation built wrongly. Then ask whether the answer can happen: a length must be positive, and a count of batches must be a whole number.

Worked examples

Examples 1 to 4 go with Part 1. Examples 5 to 7 go with Part 2.

Example 1

Combining like terms first

Solve (a) 7a - 2 + 3a + 9 = 47 \qquad (b) 18 = 2m + 11 - 5m + 1

Solution

(a) The left side has two terms in a and two constants. Reorder them (commutative property) and combine: 7a + 3a = 10a and -2 + 9 = 7. Then subtract 7 from both sides and divide both sides by 10:

\begin{aligned} 10a + 7 &= 47 \\ 10a &= 40 \\ a &= 4 \end{aligned}

Check in the original: 7(4) - 2 + 3(4) + 9 is 28 - 2 + 12 + 9 = 47.

(b) The unknown is on the right, which changes nothing. Combine 2m - 5m = -3m and 11 + 1 = 12, so the right side is -3m + 12. Subtract 12 from both sides, then divide both sides by -3:

\begin{aligned} 18 &= -3m + 12 \\ 6 &= -3m \\ -2 &= m \end{aligned}

So m = -2. Check: 2(-2) + 11 - 5(-2) + 1 is -4 + 11 + 10 + 1 = 18.

Example 2

Brackets, with a minus sign in front

Solve (a) 3(2k + 5) - 4k = 21 \qquad (b) 10 - 2(n - 6) = 4

Solution

(a) Distribute the 3 to both terms in the bracket, then combine 6k - 4k = 2k:

\begin{aligned} 6k + 15 - 4k &= 21 \\ 2k + 15 &= 21 \\ 2k &= 6 \\ k &= 3 \end{aligned}

Check: 3(2 \cdot 3 + 5) - 4(3) is 33 - 12 = 21.

(b) The term is -2(n - 6), so the number multiplying the bracket is -2, sign included. Then (-2)(n) = -2n and (-2)(-6) = 12. Combine the constants 10 + 12 = 22:

\begin{aligned} 10 - 2n + 12 &= 4 \\ 22 - 2n &= 4 \\ -2n &= -18 \\ n &= 9 \end{aligned}

Check: 10 - 2(9 - 6) = 10 - 6 = 4.

\begin{aligned} 6w - 2 - w - 4 &= 9 \\ 5w - 6 &= 9 \\ 5w &= 15 \\ w &= 3 \end{aligned}

Check: 2(3 \cdot 3 - 1) - (3 + 4) is 16 - 7 = 9.

Example 3

A rectangle’s perimeter

Ugo cuts a rectangle of card whose length is (3x - 2) cm and whose width is (x + 1) cm. Its perimeter is 46 cm. Find x and the two side lengths.

Solution

Figure 1The rectangle in Example 3. Each label appears twice, because opposite sides of a rectangle are equal.

The perimeter is told two ways: as the distance around the four sides, and as 46. A rectangle’s opposite sides are equal, so the four sides are two lengths and two widths:

2(3x - 2) + 2(x + 1) = 46

Distribute each 2, then combine 6x + 2x = 8x and -4 + 2 = -2:

\begin{aligned} 6x - 4 + 2x + 2 &= 46 \\ 8x - 2 &= 46 \\ 8x &= 48 \\ x &= 6 \end{aligned}

The length is 3(6) - 2 = 16 cm and the width is 6 + 1 = 7 cm.

Check in the story. Around the card: 16 + 7 + 16 + 7 = 46 cm. Both sides are positive, as lengths must be.

Example 4

Consecutive integers

(a) Three consecutive integers add up to 147. Find them.

(b) Can two consecutive integers add up to 100?

Solution

Consecutive integers follow each other with no gap, such as 8, 9, 10 or -3, -2, -1. Call the smallest one n. The next is one more, n + 1, and the one after that is n + 2.

Figure 2Three consecutive integers. Each is one more than the one before, so all three can be written with the one letter n.

(a) The sum is told two ways:

n + (n + 1) + (n + 2) = 147

The brackets only group terms that are being added, so they can be dropped (associative property). Combine the three n terms and the constants 1 + 2 = 3:

\begin{aligned} 3n + 3 &= 147 \\ 3n &= 144 \\ n &= 48 \end{aligned}

The integers are 48, 49 and 50. Check: 48 + 49 + 50 = 147.

(b) With n the smaller, n + (n + 1) = 100 tidies to 2n + 1 = 100, so 2n = 99 and \displaystyle n = \frac{99}{2}. The equation has a solution, but it is not an integer, so no: no two consecutive integers add up to 100. The nearest sums are 49 + 50 = 99 and 50 + 51 = 101.

The tidied side explains why. 2n is even, so 2n + 1, the sum of any two consecutive integers, is always odd, and 100 is even.

Example 5

Clearing fractions

Solve (a) \displaystyle \frac{1}{2}(y - 5) + \frac{2}{3}y = 1 \qquad (b) \displaystyle \frac{2m + 1}{5} - \frac{m}{2} = 1

Solution

(a) The denominators are 2 and 3, with least common multiple 6. Multiply every term of both sides by 6: \displaystyle 6 \cdot \frac{1}{2}(y - 5) = 3(y - 5) (associative property), \displaystyle 6 \cdot \frac{2}{3}y = 4y, and 6 \cdot 1 = 6. Then distribute and combine:

\begin{aligned} 3(y - 5) + 4y &= 6 \\ 3y - 15 + 4y &= 6 \\ 7y - 15 &= 6 \\ 7y &= 21 \\ y &= 3 \end{aligned}

Check: \displaystyle \frac{1}{2}(3 - 5) + \frac{2}{3}(3) = -1 + 2 = 1.

(b) The denominators are 5 and 2, with least common multiple 10. A fraction bar groups its numerator: \displaystyle \frac{2m + 1}{5} means (2m + 1) \div 5. So \displaystyle 10 \cdot \frac{2m + 1}{5} = 2(2m + 1), a bracket that keeps the 2 attached to both terms, and \displaystyle 10 \cdot \frac{m}{2} = 5m:

\begin{aligned} 2(2m + 1) - 5m &= 10 \\ 4m + 2 - 5m &= 10 \\ -m + 2 &= 10 \\ -m &= 8 \\ m &= -8 \end{aligned}

Check: \displaystyle \frac{2(-8) + 1}{5} - \frac{-8}{2} is -3 + 4 = 1.

Example 6

Clearing decimals

Solve (a) 1.2a - 0.35 + 0.8a = 2.65 \qquad (b) 0.25(x + 8) - 0.1x = 3.5

Solution

(a) The decimals with the most places, 0.35 and 2.65, have two, so multiply every term by 100:

\begin{aligned} 120a - 35 + 80a &= 265 \\ 200a - 35 &= 265 \\ 200a &= 300 \\ a &= 1.5 \end{aligned}

Check: 1.2(1.5) - 0.35 + 0.8(1.5) is 1.8 - 0.35 + 1.2 = 2.65. Clearing is a choice, not a rule: here combining first gives 2a - 0.35 = 2.65, just as easy.

(b) The most places is two, in 0.25, so multiply by 100. In 0.25(x + 8) only the 0.25 is a decimal, and by the associative property of multiplication 100 \cdot 0.25(x + 8) = 25(x + 8): the 100 multiplies the 0.25 and the bracket waits to be distributed. Then 100 \cdot 0.1x = 10x and 100 \cdot 3.5 = 350:

\begin{aligned} 25(x + 8) - 10x &= 350 \\ 25x + 200 - 10x &= 350 \\ 15x + 200 &= 350 \\ 15x &= 150 \\ x &= 10 \end{aligned}

Check: 0.25(10 + 8) - 0.1(10) = 4.5 - 1 = 3.5.

Example 7

A repair bill

Quentin-Ash’s phone repair bill lists a new screen at 64.90 dollars, a new battery at 27.35 dollars, and labor at 42.50 dollars an hour. The total is 156.00 dollars. How many hours of labor were charged?

Solution

Let h be the number of hours. The total is told two ways: built from the bill it is the two parts plus 42.50 dollars for each of h hours, and it is 156 dollars:

64.9 + 27.35 + 42.5h = 156

Combine the two constants, 64.90 + 27.35 = 92.25. Then multiply every term by 100 to clear the decimals, subtract, and divide:

\begin{aligned} 92.25 + 42.5h &= 156 \\ 9225 + 4250h &= 15600 \\ 4250h &= 6375 \\ h &= 1.5 \end{aligned}

Check in the story. One and a half hours at 42.50 dollars is 63.75 dollars, and 64.90 + 27.35 + 63.75 = 156.00. So 1.5 hours of labor. A time need not be a whole number of hours.

Boundaries

The unknown is on one side only. Every equation here has its letter on one side, so tidying that side reaches the two-step shape. When the unknown is on both sides, as in 7x - 4 = 3x + 8, one more move is needed first: gathering the unknown onto one side.

The unknown must survive the tidying. If the terms in x cancel when combined, there is no two-step equation left. The equation 3(x + 2) - 3x = 6 tidies to 6 = 6, true for every number, and 3(x + 2) - 3x = 5 tidies to 6 = 5, true for none. Those cases are no solution, or every number.

Clearing multiplies every term of both sides, by a number that is not 0. Multiplying only the fractions, and not the whole-number terms, changes the equation. Multiplying by 0 would turn any equation into 0 = 0 and lose all its information, which is why the multiplication property of equality needs a non-zero number.

A power of ten clears only terminating decimals. A repeating decimal such as 0.333\ldots has no last digit, so no power of ten makes it whole. Write it as the fraction \displaystyle \frac{1}{3} and clear that instead.

Letters only in numerators. An unknown in a denominator, as in \displaystyle \frac{9}{x} - 1 = 2, is a later topic: solving rational equations.

A story can reject an answer the equation accepts. A length must be positive, and a count of integers, people or batches must be a whole number. In Example 4(b) the equation’s solution is \displaystyle \frac{99}{2}, so no integers fit.

Where this goes next

Next comes equations with the unknown on both sides: one subtraction gathers the letter on one side, and then the work of this lesson finishes the job. Together with equations with no solution or every number as a solution that covers every linear equation in one unknown.

The same tidying, done with letters standing for numbers, is solving a formula for one of its letters. With the equals sign replaced by < or >, the steps carry over to multi-step inequalities, with one extra rule when multiplying or dividing by a negative. And a story like the repair bill, a fixed amount plus a rate times a quantity, is a linear model: this lesson found one point on its line.

Exercises

Part 1: combining like terms

Solve each equation and check your answer in the original.

4x + 7 + 2x - 3 = 34
9 - 3y + 8y = -16
25 = 6a - 4 - 2a + 1
-m + 12 - 4m = 2

Part 1: brackets

5(p + 2) - 3p = 24
7 - 3(t - 4) = 31
4(2r - 1) - (3r + 6) = 15
6(z + 1) + 2(z - 7) = 12
-2(3 - c) + 5(c - 1) = 3

Part 1: from words

Write an equation, solve it, and check the answer in the story.

A rectangle has length (2x + 1) m and width (x - 3) m. Its perimeter is 50 m. Find x and the length and width.
Three consecutive integers add up to -36. Find them.
Three consecutive even integers add up to 78. Find them.
Stellan’s piano tuner charges 85 dollars for parts, a 15-dollar travel fee, and labor at 40 dollars an hour. The total is 200 dollars. How many hours of labor were charged?

Part 2: clearing fractions

\displaystyle \frac{x}{2} + \frac{x}{5} = 14
\displaystyle \frac{2}{3}k - \frac{1}{6}k + 1 = 4
\displaystyle \frac{1}{3}(w + 4) - \frac{w}{2} = 1
\displaystyle \frac{a - 3}{4} + \frac{a}{6} = 2

Part 2: clearing decimals

0.3x + 0.45x = 6
0.2(n - 5) + 0.15n = 2.5
1.5 - 0.4(y + 2) = -0.5

Part 2: from words

Write an equation, solve it, and check the answer in the story.

Verity-Anne’s bicycle repair bill lists a chain at 36.80 dollars, a tire at 12.45 dollars, and labor at 38 dollars an hour. The total is 106.25 dollars. How many hours of labor were charged?
Tamsin-Rae bakes several batches of the same cookies. Each batch uses \displaystyle \frac{2}{3} of a cup of sugar in the dough and \displaystyle \frac{1}{4} of a cup in the icing. She uses \displaystyle 5\frac{1}{2} cups of sugar in all. How many batches does she bake?
Three friends share a bill. A coupon takes one fifth off the bill, and they split the rest equally, each paying 16 dollars. What was the bill before the coupon?

Reasoning

Clear the fractions in \displaystyle \frac{x}{2} + \frac{x}{5} = 14 by multiplying both sides by 20 instead of 10. Show that the answer is the same, and explain why any common multiple of the denominators works.
Consecutive odd integers go up by 2, as in 7, 9, 11. Find three consecutive odd integers that add up to 81. Then show from the equation that no three add up to 80, and give a second reason that does not use the equation.

Answers

x = 5. Combine: 6x + 4 = 34, so 6x = 30. Check: 20 + 7 + 10 - 3 = 34.
y = -5. Combine: 5y + 9 = -16, so 5y = -25. Check: 9 + 15 - 40 = -16.
a = 7. Combine: 25 = 4a - 3, so 28 = 4a. Check: 42 - 4 - 14 + 1 = 25.
m = 2. Combine: -5m + 12 = 2, so -5m = -10. Check: -2 + 12 - 8 = 2.
p = 7. Distribute: 5p + 10 - 3p = 24, so 2p + 10 = 24 and 2p = 14.
t = -4. Distribute -3: 7 - 3t + 12 = 31, so 19 - 3t = 31 and -3t = 12. Check: 7 - 3(-8) = 7 + 24 = 31.
r = 5. Distribute: 8r - 4 - 3r - 6 = 15, so 5r - 10 = 15 and 5r = 25. Check: 4(9) - 21 = 15.
\displaystyle z = \frac{5}{2}. Distribute: 6z + 6 + 2z - 14 = 12, so 8z - 8 = 12 and 8z = 20. Check: 6(3.5) + 2(-4.5) = 21 - 9 = 12.
c = 2. Distribute: -6 + 2c + 5c - 5 = 3, so 7c - 11 = 3 and 7c = 14. Check: -2(1) + 5(1) = 3.
x = 9; the length is 19 m and the width 6 m. 2(2x + 1) + 2(x - 3) = 50 gives 4x + 2 + 2x - 6 = 50, so 6x - 4 = 50 and 6x = 54. Check: 19 + 6 + 19 + 6 = 50.
-13, -12 and -11. n + (n + 1) + (n + 2) = -36 gives 3n + 3 = -36, so 3n = -39 and n = -13. Check: -13 - 12 - 11 = -36.
24, 26 and 28. Consecutive even integers go up by 2: n + (n + 2) + (n + 4) = 78 gives 3n + 6 = 78, so 3n = 72 and n = 24. Check: 24 + 26 + 28 = 78.
2.5 hours. 85 + 15 + 40h = 200 gives 100 + 40h = 200, so 40h = 100. Check: 40 \times 2.5 = 100, and 85 + 15 + 100 = 200.
x = 20. Multiply by 10: 5x + 2x = 140, so 7x = 140. Check: 10 + 4 = 14.
k = 6. Multiply by 6: 4k - k + 6 = 24, so 3k = 18. Check: 4 - 1 + 1 = 4.
w = 2. Multiply by 6: 2(w + 4) - 3w = 6, so 2w + 8 - 3w = 6 and -w = -2. Check: \displaystyle \frac{1}{3}(6) - 1 = 2 - 1 = 1.
\displaystyle a = \frac{33}{5}. Multiply by 12: 3(a - 3) + 2a = 24, so 5a - 9 = 24 and 5a = 33. Check: \displaystyle a - 3 = \frac{18}{5}, so the terms are \displaystyle \frac{9}{10} and \displaystyle \frac{11}{10}, which add to 2.
x = 8. Multiply by 100: 30x + 45x = 600, so 75x = 600. Check: 2.4 + 3.6 = 6.
n = 10. Multiply by 100: 20(n - 5) + 15n = 250, so 35n - 100 = 250 and 35n = 350. Check: 0.2(5) + 1.5 = 2.5.
y = 3. Multiply by 10: 15 - 4(y + 2) = -5, so 7 - 4y = -5 and -4y = -12. Check: 1.5 - 0.4(5) = 1.5 - 2 = -0.5.
1.5 hours. 36.8 + 12.45 + 38h = 106.25 gives 49.25 + 38h = 106.25, so 38h = 57. Check: 38 \times 1.5 = 57, and 36.80 + 12.45 + 57 = 106.25.
6 batches. \displaystyle \frac{2}{3}b + \frac{1}{4}b = \frac{11}{2}; multiply by 12: 8b + 3b = 66, so 11b = 66. Check: \displaystyle 6 \cdot \frac{2}{3} = 4 cups and \displaystyle 6 \cdot \frac{1}{4} = 1\frac{1}{2} cups, \displaystyle 5\frac{1}{2} in all.
60 dollars. Together they pay 3 \times 16 = 48 dollars, so \displaystyle b - \frac{1}{5}b = 48. Multiply by 5: 5b - b = 240, so 4b = 240. Check: one fifth of 60 is 12, leaving 48, and 48 \div 3 = 16.
x = 20 again. Multiplying by 20 gives 10x + 4x = 280, so 14x = 280. A common multiple M is not 0, so multiplying by it keeps the same solutions; and since each denominator divides M, every fraction becomes a whole-number multiple of x. A larger M only makes the numbers bigger.
25, 27 and 29. With n the smallest, n + (n + 2) + (n + 4) = 81 tidies to 3n + 6 = 81, so 3n = 75 and n = 25, which is odd. Check: 25 + 27 + 29 = 81. For 80, 3n + 6 = 80 gives 3n = 74 and \displaystyle n = \frac{74}{3}, not an integer, so no three fit. Second reason: two odd numbers add to an even number, and an even number plus an odd one is odd, so three odd integers always add to an odd number, and 80 is even.