# One-Step Equations Containing Fractions

- -

After defining one–step equations and learning how to solve those if integers or decimals are involved, now let us learn how to solve one–step equations containing fractions. The rules for solving one–step equations containing integers or decimals still hold true for one–step equations involving fractions.

Still the underlying purpose is to separate the variable and the constant part, by placing the variable on the left side and the constant on the right side of the equation. To do so, perform the opposite of whatever operation is involved in the original equation or by doing the inverse of what is being done to the variable. Since fractions are involved, please note the rules for performing operations with fractions.

Listed below are some examples of one–step equations containing fractions:

1. $a+\displaystyle \frac{3}{5}=\displaystyle \frac{1}{5}$
2. $b-\displaystyle \frac{3}{9}=\displaystyle \frac{2}{3}$
3. $w+\displaystyle \frac{1}{4}=6\displaystyle \frac{3}{4}$
4. $3w=2\displaystyle \frac{4}{10}$
5. $\displaystyle \frac{3}{100}a=24$
6. $\displaystyle \frac{1}{2}h=\displaystyle \frac{1}{5}$

Solving One – step Equations Containing Fractions

To solve one–step equation, do the inverse of the operation involved. Since, we are dealing with equations, it must be noted that whatever operation is done on one side of the equation must as well be done on the other side of the equation. Also, apply the rules for performing operations with fractions. To illustrate, consider the following examples:

Solve for the unknown variable:

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

Solution:

To solve for $a$, we subtract $\displaystyle \frac{1}{5}$ to both sides of the equation.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &a+\displaystyle \frac{1}{5}&=&1\displaystyle \frac{2}{5} \\ &a+\displaystyle \frac{1}{5}-\displaystyle \frac{1}{5}&=&1\displaystyle \frac{2}{5}-\displaystyle \frac{1}{5} \\ &\therefore a&=&1\displaystyle \frac{1}{5} \\ \end{array}$

2. $b-2 \displaystyle \frac{3}{10}=3\displaystyle \frac{4}{10}$

Solution:

To solve for $b$, add $2 \displaystyle \frac{3}{10}$ to both sides then simplify the fractions on the right side of the equation.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &b-2\displaystyle \frac{3}{10}&=&3\displaystyle \frac{4}{10} \\ &b-2\displaystyle \frac{3}{10}+2\displaystyle \frac{3}{10}&=&3\displaystyle \frac{4}{10}+2\displaystyle \frac{3}{10} \\ &\therefore b&=&5\displaystyle \frac{7}{10} \\ \end{array}$

3. $3w=2\displaystyle \frac{4}{10}$

Solution:

To solve for $w$, multiply both sides by $\displaystyle \frac{1}{3}$, then simplify the right–hand side of the equation.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &3w&=&2\displaystyle \frac{4}{10} \\ &\left (\displaystyle \frac{1}{3} \right )3w&=&2\displaystyle \frac{4}{10}\left (\displaystyle \frac{1}{3} \right ) \\ &w&=&\displaystyle \frac{24}{10}\left (\displaystyle \frac{1}{3} \right ) \\ &\therefore w&=&\displaystyle \frac{4}{5} \\ \end{array}$

4. $\displaystyle \frac{h}{\displaystyle \frac{1}{2}}=2$

Solution:

To solve for $h$, multiply both sides by $\frac{1}{2}$ then simplify the right–hand side.

$\begin{array}{c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c} &\displaystyle \frac{h}{\displaystyle \frac{1}{2}}&=&2 \\ &\left (\displaystyle \frac{1}{2} \right )\displaystyle \frac{h}{\displaystyle \frac{1}{2}}&=&2\left (\displaystyle \frac{1}{2} \right ) \\ &\therefore h&=&1 \\ \end{array}$

Practice Exercises:

1. $a-\displaystyle \frac{3}{21} =1\displaystyle \frac{1}{25}$
2. $b+1\displaystyle \frac{3}{10} =2\displaystyle \frac{3}{10}$
3. $2c =12\displaystyle \frac{2}{5}$
4. $\displaystyle \frac{1}{5}d =26\displaystyle \frac{4}{5}$
5. $\displaystyle \frac{e}{1\displaystyle \frac{1}{5}} =14\displaystyle \frac{2}{5}$