Solve Equations with Fraction or Decimal Coefficients

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation $\frac{1}{8} x + \frac{1}{2} = \frac{1}{4} .$

	$A mathematical equation is displayed: 1/8x + 1/2 = 1/4. This is a linear equation with one variable, x, involving fractions.$
To isolate the $x$ term, subtract $\frac{1}{2}$ from both sides.
Simplify the left side.
Change the constants to equivalent fractions with the LCD.	$A mathematical equation is displayed: 1/8x = 1/4 - 2/4. The equation involves fractions and a variable 'x'.$
Subtract.
Multiply both sides by the reciprocal of $\frac{1}{8}$ .
Simplify.

This method worked fine, but many students don’t feel very confident when they see all those fractions. So we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions. Let’s solve the same equation again, but this time use the method that clears the fractions.

Solve: $\frac{1}{8} x + \frac{1}{2} = \frac{1}{4} .$

Solution

Find the least common denominator of all the fractions in the equation.
Multiply both sides of the equation by that LCD, 8. This clears the fractions.
Use the Distributive Property.
Simplify — and notice, no more fractions!
Solve using the General Strategy for Solving Linear Equations.
Simplify.
Check: Let $x = −2$

Notice in Example 1 that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.

Solve: $7 = \frac{1}{2} x + \frac{3}{4} x - \frac{2}{3} x .$

Solution

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Find the least common denominator of all the fractions in the equation.	$A mathematical equation shows '7 = (1/2)x + (3/4)x - (2/3)x' followed by 'LCD = 12', indicating the least common denominator for the fractions in the expression.$
Multiply both sides of the equation by 12.
Distribute.
Simplify — and notice, no more fractions!
Combine like terms.
Divide by 7.
Simplify.
Check: Let $x = 12 .$

In the next example, we’ll have variables and fractions on both sides of the equation.

Solve: $x + \frac{1}{3} = \frac{1}{6} x - \frac{1}{2} .$

Solution

Find the LCD of all the fractions in the equation.	$A mathematical equation showing x + 1/3 = 1/6x - 1/2, with LCD = 6. This represents a linear equation with fractions where the least common denominator is 6.$
Multiply both sides by the LCD.
Distribute.	$A mathematical equation showing the multiplication of fractions and a variable 'x': 6 * x + 6 * (1/3) = 6 * (1/6) * x - 6 * (1/2).$
Simplify — no more fractions!
Subtract $x$ from both sides.
Simplify.
Subtract 2 from both sides.
Simplify.
Divide by 5.
Simplify.
Check: Substitute $x = −1 .$

In Example 4, we’ll start by using the Distributive Property. This step will clear the fractions right away!

Solve: $1 = \frac{1}{2} (4 x + 2) .$

Solution


Distribute.
Simplify. Now there are no fractions to clear!
Subtract 1 from both sides.
Simplify.
Divide by 2.
Simplify.
Check: Let $x = 0 .$

Many times, there will still be fractions, even after distributing.

Solve: $\frac{1}{2} (y - 5) = \frac{1}{4} (y - 1) .$

Solution


Distribute.
Simplify.
Multiply by the LCD, 4.
Distribute.
Simplify.
Collect the $y$ terms to the left.
Simplify.
Collect the constants to the right.
Simplify.
Check: Substitute $9$ for $y .$

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, $0.3 = \frac{3}{10}$ and $0.17 = \frac{17}{100} .$ So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator.

Solve: $0.8 x - 5 = 7 .$

Solution

The only decimal in the equation is $0.8 .$ Since $0.8 = \frac{8}{10},$ the LCD is $10 .$ We can multiply both sides by $10$ to clear the decimal.


Multiply both sides by the LCD.
Distribute.
Multiply, and notice, no more decimals!
Add 50 to get all constants to the right.
Simplify.
Divide both sides by 8.
Simplify.
Check: Let $x = 15 .$

Solve: $0.06 x + 0.02 = 0.25 x - 1.5 .$

Solution

Look at the decimals and think of the equivalent fractions.

$0.06 = \frac{6}{100}, 0.02 = \frac{2}{100}, 0.25 = \frac{25}{100}, 1.5 = 1 \frac{5}{10}$

Notice, the LCD is $100 .$

By multiplying by the LCD we will clear the decimals.


Multiply both sides by 100.
Distribute.
Multiply, and now no more decimals.
Collect the variables to the right.
Simplify.
Collect the constants to the left.
Simplify.
Divide by 19.
Simplify.
Check: Let $x = 8 .$

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

Solve: $0.25 x + 0.05 (x + 3) = 2.85 .$

Solution


Distribute first.
Combine like terms.
To clear decimals, multiply by 100.
Distribute.
Subtract 15 from both sides.
Simplify.
Divide by 30.
Simplify.
Check: Let $x = 9 .$

Key Concepts

Solve equations with fraction coefficients by clearing the fractions.
1. Find the least common denominator of all the fractions in the equation.
2. Multiply both sides of the equation by that LCD. This clears the fractions.
3. Solve using the General Strategy for Solving Linear Equations.

Section Exercises

Practice Makes Perfect

Solve equations with fraction coefficients

In the following exercises, solve the equation by clearing the fractions.

$\frac{1}{4} x - \frac{1}{2} = - \frac{3}{4}$

Solution

x = −1

$\frac{3}{4} x - \frac{1}{2} = \frac{1}{4}$

$\frac{5}{6} y - \frac{2}{3} = - \frac{3}{2}$

Solution

y = −1

$\frac{5}{6} y - \frac{1}{3} = - \frac{7}{6}$

$\frac{1}{2} a + \frac{3}{8} = \frac{3}{4}$

Solution

$a = \frac{3}{4}$

$\frac{5}{8} b + \frac{1}{2} = - \frac{3}{4}$

$2 = \frac{1}{3} x - \frac{1}{2} x + \frac{2}{3} x$

Solution

x = 4

$2 = \frac{3}{5} x - \frac{1}{3} x + \frac{2}{5} x$

$\frac{1}{4} m - \frac{4}{5} m + \frac{1}{2} m = −1$

Solution

m = 20

$\frac{5}{6} n - \frac{1}{4} n - \frac{1}{2} n = −2$

$x + \frac{1}{2} = \frac{2}{3} x - \frac{1}{2}$

Solution

x = −3

$x + \frac{3}{4} = \frac{1}{2} x - \frac{5}{4}$

$\frac{1}{3} w + \frac{5}{4} = w - \frac{1}{4}$

Solution

$w = \frac{9}{4}$

$\frac{3}{2} z + \frac{1}{3} = z - \frac{2}{3}$

$\frac{1}{2} x - \frac{1}{4} = \frac{1}{12} x + \frac{1}{6}$

Solution

x = 1

$\frac{1}{2} a - \frac{1}{4} = \frac{1}{6} a + \frac{1}{12}$

$\frac{1}{3} b + \frac{1}{5} = \frac{2}{5} b - \frac{3}{5}$

Solution

b = 12

$\frac{1}{3} x + \frac{2}{5} = \frac{1}{5} x - \frac{2}{5}$

$1 = \frac{1}{6} (12 x - 6)$

Solution

x = 1

$1 = \frac{1}{5} (15 x - 10)$

$\frac{1}{4} (p - 7) = \frac{1}{3} (p + 5)$

Solution

p = −41

$\frac{1}{5} (q + 3) = \frac{1}{2} (q - 3)$

$\frac{1}{2} (x + 4) = \frac{3}{4}$

Solution

$x = - \frac{5}{2}$

$\frac{1}{3} (x + 5) = \frac{5}{6}$

Solve Equations with Decimal Coefficients

In the following exercises, solve the equation by clearing the decimals.

$0.6 y + 3 = 9$

Solution

y = 10

$0.4 y - 4 = 2$

$3.6 j - 2 = 5.2$

Solution

j = 2

$2.1 k + 3 = 7.2$

$0.4 x + 0.6 = 0.5 x - 1.2$

Solution

x = 18

$0.7 x + 0.4 = 0.6 x + 2.4$

$0.23 x + 1.47 = 0.37 x - 1.05$

Solution

x = 18

$0.48 x + 1.56 = 0.58 x - 0.64$

$0.9 x - 1.25 = 0.75 x + 1.75$

Solution

x = 20

$1.2 x - 0.91 = 0.8 x + 2.29$

$0.05 n + 0.10 (n + 8) = 2.15$

Solution

n = 9

$0.05 n + 0.10 (n + 7) = 3.55$

$0.10 d + 0.25 (d + 5) = 4.05$

Solution

d = 8

$0.10 d + 0.25 (d + 7) = 5.25$

$0.05 (q - 5) + 0.25 q = 3.05$

Solution

q = 11

$0.05 (q - 8) + 0.25 q = 4.10$

Everyday Math

Coins Taylor has $$2.00$ in dimes and pennies. The number of pennies is $2$ more than the number of dimes. Solve the equation $0.10 d + 0.01 (d + 2) = 2$ for $d,$ the number of dimes.

Solution

d = 18

Stamps Travis bought $$9.45$ worth of $49-cent$ stamps and $21-cent$ stamps. The number of $21-cent$ stamps was $5$ less than the number of $49-cent$ stamps. Solve the equation $0.49 s + 0.21 (s - 5) = 9.45$ for $s,$ to find the number of $49-cent$ stamps Travis bought.

Writing Exercises

Explain how to find the least common denominator of $\frac{3}{8}, \frac{1}{6}, and \frac{2}{3} .$

Solution

Answers will vary.

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

Solution

Answers will vary.

In the equation $0.35 x + 2.1 = 3.85,$ what is the LCD? How do you know?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A self-assessment chart for math skills, asking students to rate their ability to solve equations using a general strategy, with fraction coefficients, and with decimal coefficients, across three confidence levels.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

Chapter Review Exercises

Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, determine whether the given number is a solution to the equation.

$x + 16 = 31, x = 15$

Solution

yes

$w - 8 = 5, w = 3$

$−9 n = 45, n = 54$

Solution

$4 a = 72, a = 18$

In the following exercises, solve the equation using the Subtraction Property of Equality.

$x + 7 = 19$

Solution

$y + 2 = −6$

$a + \frac{1}{3} = \frac{5}{3}$

Solution

$a = \frac{4}{3}$

$n + 3.6 = 5.1$

In the following exercises, solve the equation using the Addition Property of Equality.

$u - 7 = 10$

Solution

u = 17

$x - 9 = −4$

$c - \frac{3}{11} = \frac{9}{11}$

Solution

$c = \frac{12}{11}$

$p - 4.8 = 14$

In the following exercises, solve the equation.

$n - 12 = 32$

Solution

n = 44

$y + 16 = −9$

$f + \frac{2}{3} = 4$

Solution

$f = \frac{10}{3}$

$d - 3.9 = 8.2$

$y + 8 - 15 = −3$

Solution

y = 4

$7 x + 10 - 6 x + 3 = 5$

$6 (n - 1) - 5 n = −14$

Solution

n = −8

$8 (3 p + 5) - 23 (p - 1) = 35$

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

The sum of $−6$ and $m$ is $25 .$

Solution

−6 + m = 25; m = 31

Four less than $n$ is $13 .$

In the following exercises, translate into an algebraic equation and solve.

Rochelle’s daughter is $11$ years old. Her son is $3$ years younger. How old is her son?

Solution

s = 11 − 3; 8 years old

Tan weighs $146$ pounds. Minh weighs $15$ pounds more than Tan. How much does Minh weigh?

Peter paid $$9.75$ to go to the movies, which was $$46.25$ less than he paid to go to a concert. How much did he pay for the concert?

Solution

c − 46.25 = 9.75; $56.00

Elissa earned $$152.84$ this week, which was $$21.65$ more than she earned last week. How much did she earn last week?

Solve Equations using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the Division Property of Equality.

$8 x = 72$

Solution

x = 9

$13 a = −65$

$0.25 p = 5.25$

Solution

p = 21

$- y = 4$

In the following exercises, solve each equation using the Multiplication Property of Equality.

$\frac{n}{6} = 18$

Solution

n = 108

$\frac{y}{−10} = 30$

$36 = \frac{3}{4} x$

Solution

x = 48

$\frac{5}{8} u = \frac{15}{16}$

In the following exercises, solve each equation.

$−18 m = −72$

Solution

m = 4

$\frac{c}{9} = 36$

$0.45 x = 6.75$

Solution

x = 15

$\frac{11}{12} = \frac{2}{3} y$

$5 r - 3 r + 9 r = 35 - 2$

Solution

r = 3

$24 x + 8 x - 11 x = −7 −14$

Solve Equations with Variables and Constants on Both Sides

In the following exercises, solve the equations with constants on both sides.

$8 p + 7 = 47$

Solution

p = 5

$10 w - 5 = 65$

$3 x + 19 = −47$

Solution

x = −22

$32 = −4 - 9 n$

In the following exercises, solve the equations with variables on both sides.

$7 y = 6 y - 13$

Solution

y = −13

$5 a + 21 = 2 a$

$k = −6 k - 35$

Solution

k = −5

$4 x - \frac{3}{8} = 3 x$

In the following exercises, solve the equations with constants and variables on both sides.

$12 x - 9 = 3 x + 45$

Solution

x = 6

$5 n - 20 = −7 n - 80$

$4 u + 16 = −19 - u$

Solution

u = −7

$\frac{5}{8} c - 4 = \frac{3}{8} c + 4$

In the following exercises, solve each linear equation using the general strategy.

$6 (x + 6) = 24$

Solution

x = −2

$9 (2 p - 5) = 72$

$- (s + 4) = 18$

Solution

s = −22

$8 + 3 (n - 9) = 17$

$23 - 3 (y - 7) = 8$

Solution

y = 12

$\frac{1}{3} (6 m + 21) = m - 7$

$8 (r - 2) = 6 (r + 10)$

Solution

r = 38

$5 + 7 (2 - 5 x) = 2 (9 x + 1) - (13 x - 57)$

$4 (3.5 y + 0.25) = 365$

Solution

y = 26

$0.25 (q - 8) = 0.1 (q + 7)$

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation by clearing the fractions.

$\frac{2}{5} n - \frac{1}{10} = \frac{7}{10}$

Solution

n = 2

$\frac{1}{3} x + \frac{1}{5} x = 8$

$\frac{3}{4} a - \frac{1}{3} = \frac{1}{2} a + \frac{5}{6}$

Solution

$a = \frac{14}{3}$

$\frac{1}{2} (k + 3) = \frac{1}{3} (k + 16)$

In the following exercises, solve each equation by clearing the decimals.

$0.8 x - 0.3 = 0.7 x + 0.2$

Solution

x = 5

$0.36 u + 2.55 = 0.41 u + 6.8$

$0.6 p - 1.9 = 0.78 p + 1.7$

Solution

p = −20

$0.10 d + 0.05 (d - 4) = 2.05$

Chapter Practice Test

Determine whether each number is a solution to the equation.

3 x + 5 = 23 .

ⓐ $6$
ⓑ $\frac{23}{5}$

Solution

ⓐ yes
ⓑ no

In the following exercises, solve each equation.

$n - 18 = 31$

$9 c = 144$

Solution

c = 16

$4 y - 8 = 16$

$−8 x - 15 + 9 x - 1 = −21$

Solution

x = −5

$−15 a = 120$

$\frac{2}{3} x = 6$

Solution

x = 9

$x + 3.8 = 8.2$

$10 y = −5 y + 60$

Solution

y = 4

$8 n + 2 = 6 n + 12$

$9 m - 2 - 4 m + m = 42 - 8$

Solution

m = 6

$−5 (2 x + 1) = 45$

$- (d + 9) = 23$

Solution

d = −32

$\frac{1}{3} (6 m + 21) = m - 7$

$2 (6 x + 5) - 8 = −22$

Solution

x = −2

$8 (3 a + 5) - 7 (4 a - 3) = 20 - 3 a$

$\frac{1}{4} p + \frac{1}{3} = \frac{1}{2}$

Solution

$p = \frac{2}{3}$

$0.1 d + 0.25 (d + 8) = 4.1$

Translate and solve: The difference of twice $x$ and $4$ is $16 .$

Solution

2x − 4 = 16; x = 10

Samuel paid $$25.82$ for gas this week, which was $$3.47$ less than he paid last week. How much did he pay last week?