# 2.4.3: Multi-Step Equations with Fractions

- Page ID
- 4365

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Solve Multi-Step Equations Involving Fractions

Kathleen is getting in shape for the “Relay for Life” walk. She tells her brother, Mack, that on Monday she walked 4 miles and on Tuesday one-third as many miles as she walked on Wednesday, for a total of 24 miles. How can Mack create an equation to figure out how far his sister walked on each day?

In this concept, you will learn to solve multi-step equations involving fractions.

**Multi Step Equations with Fractions**

Before you begin to solve equations involving fractions, you may have to review the rules for operations with fractions. Here are a few tips on how to perform operations with fractions:

- To add or subtract fractions they must have the same
**denominator**. - To multiply fractions you multiply the numerators and write the product over the product of the denominators.
- To divide fractions you multiply the first
**fraction**by the reciprocal of the fraction after the division sign.

Let’s look at a problem with fractions before you begin solving equations involving fractions.

((1/4)+(1/2)−(2/3))((3/4)÷(1/2))

First, perform the addition and subtraction in the first set of parenthesis. The **least common denominator** is 12.

[(1/4)(3/3)+(1/2)(6/6)−(2/3)(4/4)]

Next, multiply each fraction by multiplying the numerators and the denominators.

((3/12)+(6/12)−(8/12))

Next, express the fractions as a single fraction.

((3+6−8)/12)

Then, do the addition and subtraction in the numerator.

(1/12)

This answer is 1/12.

Next, perform the division in the second set of parenthesis.

((3/4)÷(1/2))

First, express the division as the first fraction multiplied by the reciprocal of the second fraction.

((3/4)×(2/1))

Then, multiply the numerators and multiply the denominators.

(6/4)

This answer is (6/4).

Next, multiply the answers from each set of parenthesis.

(1/12)(6/4)

Remember, multiply the numerators and multiply the denominators.

6/48

Then, simplify the answer.

6/48=1/8

The answer is 1/8.

Let’s look at solving an equation involving fractions.

Solve the following equation for ‘x’.

x−(x/2)−(1/12)=5/6

First, isolate the variable ‘x’ by adding 112 to both sides of the equation.

x−(x/2)−(1/12)+(1/12)=(5/6)+(1/12)

Next, simplify both sides of the equation.

x−(x/2)=(10/12)+(1/12)

x−(x/2)=11/12

Next, simplify the left side of the equation by performing the subtraction.

(x/1)(2/2)−(x/2)=11/12

(2x/2)−(x/2)=11/12

1x/2=11/12

Next, multiply both sides of the equation by 12.

6x=11

Then, divide both sides of the equation by 6.

x=1(5/6)

The answer is x=1(5/6).

Some equations involving fractions will require you to apply the distributive property to clear parenthesis as you solve the equation.

Solve the following equation:

(2/3)(y+(3/5))=2

First, clear the parenthesis by applying the distributive property.

(2/3)(y/1)+(2/3)(3/5)=2

(2y/3)+(6/15)=2

Next, isolate the variable by subtracting 615 from both sides of the equation.

(2y/3)+(6/15)−(6/15)=2−(6/15)

Next, simplify each side of the equation.

2y/3=2−(6/15)

2y/3=(2/1)(15/15)−(6/15)

2y/3=(30/15)−(6/15)

2y/3=24/15

Next, multiply both sides of the equation by 15.

10y=24

Then, divide both sides of the equation by 10.

y=24/10

y=2(4/10)

y=2(2/5)

The answer is y=2(2/5).

### Examples

Example 2.4.3.1

Earlier, you were given a problem about Kathleen who was walking for “Relay for Life?”

Her brother, Mack, wants to write an equation to figure out how many miles she walked each day.

**Solution**

First, name the variable. Let ‘w’ represent the number of miles Kathleen walked on Wednesday.

Next, represent “* On Tuesday Kathleen walked one-third as many miles as… on Wednesday*.”

13w

Next, represent the information given in the problem using a verbal model.

Verbal Model:

(miles walked Mon.)+(miles walked Tues.)+(miles walked Wed.)=(total miles walked)

Next, write an equation to represent the verbal model.

4+(1/3)w+w=24

Next, solve the equation for the variable.

First, simplify the left side of the equation.

4+(1/3)w+w=24

4+1(1/3)w=24

4+(4/3)w=24

Next, isolate the variable by subtracting 4 from both sides of the equation.

4+(4/3)w=24

4−4+(4/3)w=24−4

Next, simplify both sides of the equation.

4−4+(4/3)w=24-4

4+(4/3)w=24

(4/3)w=20

Next, divide both sides of the equation by 4/3 to solve for ‘w’.

(4/3)w=20

(4/3)÷(4/3)w=20÷(4/3)

(4/3)×(3/4)w=20×(3/2)

w=60/4

w=15

The answer is w=15.

Kathleen walked 15 miles on Wednesday.

Then, use the answer to figure the number of miles Kathleen walked on Tuesday.

w=15

miles walked on Tuesday=(1/3)w

(1/3)(15)=5

The answer is 5.

Kathleen walked 5 miles on Tuesday.

Example 2.3.4.2

Solve the following equation for the variable:

(1/3)+(4/5)−n=(2/15)

**Solution**

First, add the fractions on the left side of the equation. Remember the fractions must have a common denominator.

(1/3)+(4/5)−n=2/15

(1/3)(5/5)+(4/5)(3/3)−n=2/15

(5/15)+(12/15)−n=2/15

(17/15)−n=2/15

Next, isolate the variable by subtracting 1715 from both sides of the equation.

(17/15)−n=(2/15)

(17/15)−(17/15)−n=(2/15)−(17/15)

Next, simplify both sides of the equation.

(17/15)−(17/15)−n=(2/15)−(17/15)

−n=−15/15

−n=−1

Then, divide both sides of the equation by -1 to solve for ‘n’. When you solve an equation you are finding the value of +1n.

The answer is n=1.

Example 2.3.4.3

Solve the following equation for the variable:

(3/6)−(1/3)+x=1(1/2)

**Solution**

First, subtract the fractions on the left side of the equation. Remember the fractions must have a common denominator.

(3/6)−(1/3)+x=1(1/2)

(3/6)−(1/3)(2/2)+x=1(1/2)

(3/6)−(2/6)+x=1(1/2)

(1/6)+x=1(1/2)

Next, express the mixed number 1(1/2) as an improper fraction.

(1/6)+x=(3/2)

Next isolate the variable by subtracting 16 from both sides of the equation.

(1/6)+x=(3/2)

(1/6)−(1/6)+x=(3/2)−(1/6)

Next, simplify both sides of the equation.

(1/6)−(1/6)+x=(3/2)−(1/6)

x=(3/2)(3/3)−(1/6)

x=(9/6)−(1/6)

x=(8/6)

Then, express the answer in simplest form.

x=1(2/6)

x=1(1/3)

The answer is x=1(1/3).

Example 2.3.4.4

Solve the following equation for the variable:

(2/5)(m+4)=6

**Solution**

First, apply the distributive property to clear the parenthesis.

(2/5)(m+4)=6

((2/5)⋅(m/1)+(2/5)⋅(4/1))=6

(2m/5)+(8/5)=6

Next, isolate the variable by subtracting 85 from both sides of the equation.

(2m/5)+(8/5)=6

(2m/5)+(8/5)−(8/5)=6−(8/5)

Next, simplify both sides of the equation.

(2m/5)+(8/5)−(8/5)=6−(8/5)

(2m/5)=(6/1)(5/5)−(8/5)

(2m/5)=(30/5)−(8/5)

(2m/5)=(22/5)

Next, multiply both sides of the equation by 5.

(2m/5)=22/5

(5/1)(2m/5)=(5/1)(22/5)

2m=22

Then, divide both sides of the equation by 2 to solve for ‘m’.

2m=22

2m/2=22/2

m=11

The answer is m=11.

Example 2.3.4.5

Solve the following equation for the variable.

(3/4)y−(2/5)=(1/2)

**Solution**

First, isolate the variable by adding 25 to both sides of the equation.

(3/4)y−(2/5)=1/2

(3/4)y−(2/5)+(2/5)=(1/2)+(2/5)

Next, simplify both sides of the equation.

(3/4)y−(2/5)+(2/5)=(1/2)+(2/5)

(3/4)y=(1/2)⋅(5/5)+(2/5)⋅(2/2)

(3/4)y=(5/10)+(4/10)

(3/4)y=(9/10)

Next, divide both sides of the equation by 34 to solve for ‘y’.

(3/4)y=(9/10)

(3/4)÷(3/4)y=(9/10)÷(3/4)

(3/4)×(4/3)y=(9/10)×(4/3)

y=(36/30)

y=1(6/30)

Then simplify the answer.

y=1(1/5)

The answer is y=1(1/5).

### Review

Solve the following equations:

- (1/3)x=9
- (1/2)x+(1/3)x=10
- (3/5)y+1=7
- (3/4)x=6
- (1/3)+(4/6)−x=(1/2)
- (4/7)+(2/7)−x=(2/7)
- (5/8)x=10
- (1/4)y+7=31
- (1/3)a−4=12
- (6/7)−(2/7)+x=1(17)
- (4/5)y−(3/5)y=10
- (2/3)x=8
- (5/6)−x=−(16)
- (3/4)y=(34)
- (6/8)−(2/3)+x=(1/3)

### Review (Answers)

To see the Review answers, open this PDF file and look for section 3.10.

### Vocabulary

Term | Definition |
---|---|

Decimal |
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths). |

Denominator |
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. 58 has denominator 8. |

fraction |
A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a .rational number |

Integer |
The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3... |

Least Common Denominator |
The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators. |

rational number |
A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero. |

Repeating Decimal |
A repeating decimal is a decimal number that ends with a group of digits that repeat indefinitely. 1.666... and 0.9898... are examples of repeating decimals. |

Terminating Decimal |
A terminating decimal is a decimal number that ends. The decimal number 0.25 is an example of a terminating decimal. |

### Additional Resources

Video:

Practice: **Multi-Step Equations with Fractions**