# 3.10: Solve Multi-Step Equations Involving Fractions

**At Grade**Created by: CK-12

**Practice**Equations with Fractions

Have you ever tried to figure out a problem involving mileage? Take a look at this situation.

On Sunday, Leah walked 4 miles. On Monday, Leah walked one-third as many miles as she walked on Tuesday. She walked a total of 12 miles on those 3 days.

Let \begin{align*}t\end{align*} represent the number of miles Leah walked today. Write an algebraic equation to represent the total number of miles she walked on all 3 days. Find the number of miles Leah walked on Tuesday. Find the number of miles Leah walked on Monday.

Pay attention to this Concept. It will help you to work with fractions. Then you can solve this dilemma successfully.

### Guidance

Do you know how to solve this equation that has fractions in it?

Solve for \begin{align*}n\end{align*}: \begin{align*}n - \frac{n}{2}-\frac{1}{12} = \frac{5}{6}\end{align*}

Let's look at how to do this.

**First, subtract the like terms \begin{align*}n\end{align*} and \begin{align*}\frac{n}{2}\end{align*} on the left side of the equation.** It may help to remember that \begin{align*}\frac{n}{2} = \frac{1}{2}n\end{align*} and that \begin{align*}n = 1 n = \frac{2}{2}n\end{align*}.

\begin{align*}n - \frac{n}{2} - \frac{1}{12} &= \frac{5}{6}\\ \left( \frac{2}{2}n - \frac{1}{2}n \right) - \frac{1}{12} &= \frac{5}{6}\\ \frac{n}{2} - \frac{1}{12} &= \frac{5}{6}\end{align*}

**The next step is to isolate the term with the variable, \begin{align*}\frac{n}{2}\end{align*}, on one side of the equation. Since \begin{align*}\frac{1}{12}\end{align*} is subtracted from \begin{align*}\frac{n}{2}\end{align*}, you should add \begin{align*}\frac{1}{12}\end{align*} to both sides of the equation.**

In doing this step, you will need to add \begin{align*}\frac{1}{12}\end{align*} and \begin{align*}\frac{5}{6}\end{align*}, two fractions with unlike denominators. Before you add those fractions, you will need to give them a common denominator. That means you will need to find a common multiple of those two denominators and rewrite each fraction as an equivalent fraction with that denominator. Since the least common multiple of 12 and 6 is 12, you will need to rewrite \begin{align*}\frac{5}{6}\end{align*} as an equivalent fraction with a denominator of 12. You do not need to rewrite \begin{align*}\frac{1}{12}\end{align*} since it already has a denominator of 12.

\begin{align*}\frac{n}{2}-\frac{1}{12} &= \frac{5}{6}\\ \frac{n}{2} - \frac{1}{12} + \frac{1}{12} &= \frac{5}{6} + \frac{1}{12}\\ \frac{n}{2} + \left ( - \frac{1}{12} + \frac{1}{12} \right ) &= \frac{5}{6} + \frac{1}{12} && \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}\\ \frac{n}{2} + 0 &= \frac{10}{12} + \frac{1}{12}\\ \frac{n}{2} &= \frac{11}{12}\end{align*}

**Since \begin{align*}\frac{n}{2}\end{align*} means \begin{align*}n \div 2\end{align*}, we should multiply each side of the equation by 2, or \begin{align*}\frac{2}{1}\end{align*}, to get \begin{align*}n\end{align*} by itself on one side of the equation.**

\begin{align*}\frac{n}{2} &= \frac{11}{12}\\ \frac{n}{\bcancel{2}} \times \frac{\bcancel{2}}{1} &= \frac{11}{12} \times \frac{2}{1}\\ \frac{n}{1} &= \frac{22}{12}\\ n &= \frac{11}{6} = 1 \frac{5}{6}\end{align*}

**The value of \begin{align*}n\end{align*} is \begin{align*}1 \frac{5}{6}\end{align*}.**

Some equations with fractions will also have a set of parentheses in them. To work with these problems, you will need to use the distributive property to simplify the equation.

*Solve for \begin{align*}r\end{align*}: \begin{align*}\frac{2}{3}(r + \frac{3}{5}) = 2\end{align*}*

**Apply the distributive property to the left side of the equation. Multiply each of the two numbers inside the parentheses by \begin{align*}\frac{2}{3}\end{align*} and then add those products.**

\begin{align*}\frac{2}{3} \left ( r + \frac{3}{5} \right ) &= 2\\ \left ( \frac{2}{3} \times r \right ) + \left ( \frac{2}{\bcancel{3}} \times \frac{\bcancel{3}}{5} \right ) &= 2\\ \frac{2}{3}r + \frac{2}{5} &= 2\end{align*}

**Now, solve as you would solve any two-step equation. To get the term with the variable, \begin{align*}\frac{2}{3}r\end{align*}, by itself on one side of the equation, subtract \begin{align*}\frac{2}{5}\end{align*} from both sides.** To do this, it will help to rename 2 as \begin{align*}\frac{10}{5}\end{align*}.

\begin{align*}\frac{2}{3}r + \frac{2}{5} &= 2\\ \frac{2}{3}r + \left ( \frac{2}{5} - \frac{2}{5} \right ) &= 2 - \frac{2}{5}\\ \frac{2}{3}r + 0 &= \frac{10}{5} - \frac{2}{5}\\ \frac{2}{3}r &= \frac{8}{5}\end{align*}

Since \begin{align*}\frac{2}{3}r\end{align*} means \begin{align*}\frac{2}{3} \times r\end{align*}, use the inverse of multiplication—division—and divide both sides of the equation by \begin{align*}\frac{2}{3}\end{align*}. This will involve dividing \begin{align*}\frac{2}{3}r \div \frac{2}{3}\end{align*} on the left side of the equation. Remember, to divide two fractions, take the reciprocal of the divisor (the second fraction) and multiply that reciprocal by the dividend (the first fraction). So, \begin{align*}\frac{2}{3}r \div \frac{2}{3}r \times \frac{3}{2}\end{align*}. Since you will be multiplying the left side of the equation by the reciprocal of \begin{align*}\frac{2}{3}\end{align*}, which is \begin{align*}\frac{3}{2}\end{align*}, you will need to multiply the right side of the equation by \begin{align*}\frac{3}{2}\end{align*} also.

\begin{align*}\frac{2}{3}r &= \frac{8}{5}\\ \frac{2}{3}r \div \frac{2}{3} &= \frac{8}{5} \div \frac{2}{3}\\ \frac{2}{3}r \times \frac{3}{2} &= \frac{8}{5} \times \frac{3}{2}\\ \frac{\bcancel 2}{\bcancel 3}r \times \frac{\bcancel 3}{\bcancel 2} &= \frac{24}{10}\\ 1r &= \frac{12}{5}\\ r &= 2\frac{2}{5}\end{align*}

**The value of \begin{align*}r\end{align*} is \begin{align*}2 \frac{2}{5}\end{align*}.**

Solve each for the unknown variable. Be sure your answer is in simplest form.

#### Example A

\begin{align*}\frac{1}{3} + \frac{4}{5} - n = \frac{2}{15}\end{align*}

**Solution:\begin{align*}n = 1\end{align*}**

#### Example B

\begin{align*}\frac{3}{6}-\frac{1}{3}+x=1 \frac{1}{2}\end{align*}

**Solution: \begin{align*}1 \frac{1}{3}\end{align*}**

#### Example C

\begin{align*}\frac{1}{2}+\frac{7}{8}+x = 2\end{align*}

**Solution:\begin{align*}\frac{5}{8}\end{align*}**

Now let's go back to the dilemma at the beginning of the Concept.

**Consider part** *a***first.**

You know that \begin{align*}t\end{align*} represents the number of miles Leah walked on Tuesday. Use that variable to write an expression for the number of miles Leah walked on Monday.

\begin{align*}& On \ Monday, \ Leah \ walked \ \underline{one-third \ as \ many \ miles \ as \ldots on \ Tuesday}.\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \downarrow\\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \frac{t}{3} \ or \ \frac{1}{3}t\end{align*}

So, you know that Leah walked 4 miles on Sunday, \begin{align*}t\end{align*} miles on Monday, and \begin{align*}\frac{1}{3}t\end{align*} miles on Tuesday. You also know that she walked a *total* of 12 miles on all three days. Use this information to write an addition equation for this problem.

\begin{align*}& (\text{miles walked Sun.}) + (\text{miles walked Mon.}) + (\text{miles walked Tues.}) = (\text{total miles walked})\\ & \qquad \qquad \downarrow \qquad \qquad \downarrow \qquad \qquad \ \downarrow \qquad \quad \qquad \downarrow \qquad \qquad \ \downarrow \qquad \qquad \ \ \downarrow \qquad \qquad \ \ \downarrow\\ & \qquad \qquad \ 4 \qquad \quad \ \ + \qquad \qquad \frac{1}{3}t \qquad \qquad + \qquad \qquad \ t \qquad \qquad \ = \qquad \quad \ \ 12\end{align*}

So, this problem can be represented by the equation, \begin{align*}4 + \frac{1}{3}t+t=12\end{align*}.

**Next, consider part** *b***.**

The variable \begin{align*}t\end{align*} represents the number of miles Leah walked today. So, solve the equation for \begin{align*}t\end{align*}. Start by adding the like terms on the left side of the equation.

\begin{align*}4 + \frac{1}{3}t + t &= 12\\ 4 + \frac{1}{3}t + \frac{3}{3}t &= 12\\ 4 + \frac{4}{3}t &= 12\end{align*}

Solve the equation for \begin{align*}t\end{align*} as you would solve any two-step equation. Subtract 4 from both sides of the equation.

\begin{align*}4 + \frac{4}{3}t &= 12\\ 4 - 4 + \frac{4}{3}t &= 12-4\\ 0 + \frac{4}{3}t &= 8\\ \frac{4}{3}t &= 8\end{align*}

Finally, you must divide both sides of the equation by \begin{align*}\frac{4}{3}\end{align*}. Remember, that is the same as multiplying both sides of the equation by \begin{align*}\frac{3}{4}\end{align*}.

\begin{align*}\frac{4}{3}t &= 8\\ \frac{4}{3}t \times \frac{3}{4} &= 8 \times \frac{3}{4}\\ \frac{\bcancel{4}}{\bcancel{3}}t \times \frac{\bcancel{3}}{\bcancel{4}} &= \frac{8}{1} \times \frac{3}{4}\\ 1t &= \frac{24}{4}\\ t &= 6\end{align*}

**The value of \begin{align*}t\end{align*} is 6, so Leah walked 6 miles on Tuesday.**

**Consider part** *c***next.**

In part *a*, you determined that Leah walked \begin{align*}\frac{1}{3}t\end{align*} miles on Monday. Since \begin{align*}t = 6\end{align*}, substitute 6 for \begin{align*}t\end{align*} in the expression to find how many miles she walked yesterday.

\begin{align*}\frac{1}{3}t = \frac{1}{3} \times 6 = \frac{1}{3} \times \frac{6}{1} = \frac{6}{3} = 2\end{align*}

**Leah walked 2 miles on Monday.**

### Vocabulary

- Integer
- the set of whole numbers and their opposites.

- Rational Numbers
- a set of numbers that includes integers, decimals, fractions, terminating and repeating decimals. These numbers can be written in fraction form.

- Fraction
- a part of a whole written using a numerator and a denominator.

- Decimal
- a part of a whole written using place value and a decimal point.

- Repeating Decimal
- a decimal where the digits repeat in a pattern and eventually end.

- Terminating Decimal
- a decimal where the digits eventually end, but where numbers do not repeat in a pattern.

### Guided Practice

Here is one for you to try on your own.

Solve for the unknown variable. Be sure that your answer is in simplest form.

\begin{align*}\frac{12}{13}+\frac{11}{13}-x=\frac{6}{13}\end{align*}

**Solution**

First, add the numerators of the two fractions with a common denominator.

\begin{align*}\frac{23}{13} - x = \frac{6}{13}\end{align*}

Now we have to figure out what quantity is taken away from \begin{align*}\frac{23}{13}\end{align*} to have \begin{align*}\frac{6}{13}\end{align*}.

We can convert \begin{align*}\frac{23}{13}\end{align*} to a mixed number.

\begin{align*}1 \frac{10}{13}\end{align*}

Our work is simpler now.

\begin{align*}10 - 4 = 6\end{align*}

**Our answer is \begin{align*}x = 1 \frac{4}{13}\end{align*}.**

### Video Review

Solving Two-Step Linear Equations with Fractions

### Practice

Directions: Solve each equation.

1. \begin{align*}\frac{1}{3}x = 9\end{align*}

2. \begin{align*}\frac{1}{2}x + \frac{1}{3}x = 10\end{align*}

3. \begin{align*}\frac{3}{5}y + 1 = 7\end{align*}

4. \begin{align*}\frac{3}{4}x = 6\end{align*}

5. \begin{align*}\frac{1}{3} + \frac{4}{6} - x = \frac{1}{2}\end{align*}

6. \begin{align*}\frac{4}{7}+\frac{2}{7} - x = \frac{2}{7}\end{align*}

7. \begin{align*}\frac{5}{8}x = 10\end{align*}

8. \begin{align*}\frac{1}{4}y + 7 = 31\end{align*}

9. \begin{align*}\frac{1}{3}a - 4 = 12\end{align*}

10. \begin{align*}\frac{6}{7} - {2}{7} + x = 1 \frac{1}{7}\end{align*}

11. \begin{align*}\frac{4}{5}y - \frac{3}{5}y = 10\end{align*}

12. \begin{align*}\frac{2}{3}x = 8\end{align*}

13. \begin{align*}\frac{5}{6} - x = -\frac{1}{6}\end{align*}

14. \begin{align*}\frac{3}{4}y= \frac{3}{4}\end{align*}

15. \begin{align*}\frac{6}{8} - \frac{2}{3} + x = \frac{1}{3}\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
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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. has denominator . |

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. |

### Image Attributions

Here you'll solve mult-step equations involving fractions.