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*}

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*}

Let's look at how to do this.

**First, subtract the like terms \begin{align*}n\end{align*} n and \begin{align*}\frac{n}{2}\end{align*}n2 on the left side of the equation.** It may help to remember that \begin{align*}\frac{n}{2} = \frac{1}{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*} n2, on one side of the equation. Since \begin{align*}\frac{1}{12}\end{align*}112 is subtracted from \begin{align*}\frac{n}{2}\end{align*}n2, you should add \begin{align*}\frac{1}{12}\end{align*}112 to both sides of the equation.**

In doing this step, you will need to add \begin{align*}\frac{1}{12}\end{align*}

\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*} n2 means \begin{align*}n \div 2\end{align*}n÷2, we should multiply each side of the equation by 2, or \begin{align*}\frac{2}{1}\end{align*}21, to get \begin{align*}n\end{align*}n 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*}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.

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*}\frac{1}{3}t\end{align*} miles on Monday, and \begin{align*}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*}.

Now, 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.**

How far did Leah walk on Monday?

Remember 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 on Monday.

\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*}