Solving for a Variable on One Side Using Addition and Subtraction with Fractions

Introduction

In this lesson, you will learn how to solve for a variable on one side of an equation when fractions, addition, and subtraction are involved.

This video illustrates the lesson material below. Watching the video is optional.

Solving for a Variable on One Side: Part 4 - Addition and Subtraction using Fractions (09:59 min) | Transcript

Solving for a Variable on One Side Using Addition, Subtraction, and Fractions

When solving for variables with fractions instead of whole numbers, add the additive inverse:

Additive inverse: Number when added to another number equals zero.

Example 1
\begin{align*}x + \frac{1}{2} &= 2 &\color{red}\small\text{Solve for x}\\\\x + \frac{1}{2} \color{red}\mathbf{+ (-\frac{1}{2})} &= 2 \color{red}\mathbf{+ (-\frac{1}{2})} &\color{red}\small\text{Additive inverse of $\frac{1}{2}$ is $-\frac{1}{2}$}\\\\x &= 2 - \frac{1}{2} &\color{red}\small\text{Simplifying both sides}\\\\x &= 1\frac{1}{2} &\color{red}\small\text{Subtract $\frac{1}{2}$ from 2 is $1\frac{1}{2}$}\\\\\end{align*}

If you look at the number line, you have 2, $1\frac{1}{2}$, $1$, $\frac{1}{2}$, and so on. If you start with 2 and subtract $\frac{1}{2}$, which is the same as going in the negative direction $\frac{1}{2}$, you will end up at $1\frac{1}{2}$.

Example 2
You can think of this in another way by using your skills with adding and subtracting fractions. You know that you need to find a common denominator for both fractions if you want to perform any arithmetic. In this example, you will solve for $x$ again, but this time finding a common denominator:

\begin{align*}x + \frac{1}{2} &= 2 &\color{red}\small\text{Solve for x}\\\\x + \frac{1}{2} \color{red}\mathbf{+ (-\frac{1}{2})} &= 2 \color{red}\mathbf{+ (-\frac{1}{2})} &\color{red}\small\text{Additive inverse of $\frac{1}{2}$ is $-\frac{1}{2}$}\\\\x &= \frac{2}{1} - \frac{1}{2} &\color{red}\small\text{Simplifying both sides and note: $2 = \frac{2}{1}$}\\\\x &= \frac{4}{2} - \frac{1}{2} &\color{red}\small\text{Common denominator for 1 & 2 is 2}\\\\x &= \frac{4 -1}{2} &\color{red}\small\text{Subtract the numerator}\\\\x &= \frac{3}{2} &\color{red}\small\text{Simplify the numerator: $4 -1 = 3$}\\\\\end{align*}

$\frac{3}{2}$ is the same as $1\frac{1}{2}$, so either way of solving the problem (Example 1 or 2) gives the same answer.

It’s important to note that the variable can be on either side of the equal sign. You follow the same rules for solving for the variable when x is on the right-hand side. $6=x-\frac{2}{3}$ is the same as saying $x-\frac{2}{3}=6$, so it doesn’t matter which side of the equal sign the variable is on as long as everything stays equal.

Example 3
\begin{align*}6 &= x - \frac{2}{3}&\color{red}\small\text{Solve for x}\\\\6 \color{red}\mathbf{+ (\frac{2}{3})} &= x - \frac{2}{3} \color{red}\mathbf{+ (\frac{2}{3})}&\color{red}\small\text{Additive inverse of $-\frac{2}{3}$ is $+\frac{2}{3}$}\\\\\frac{6}{1} + \frac{2}{3} &= x &\color{red}\small\text{Simplifying both sides and note: $ 6= \frac{6}{1}$}\\\\ \frac{18}{3} + \frac{2}{3} & = x&\color{red}\small\text{Common denominator for 1 & 3 is 3}\\\\\frac{18 + 2}{3}&=x &\color{red}\small\text{Add the numerator and keep the denominator}\\\\x &= \frac{20}{3} &\color{red}\small\text{Simplify the numerator: $18+2 = 20$}\\\\\end{align*}

You can check this answer by substituting $\frac{20}{3}$ for x to see if it works for the equation. It does.

Things to Remember

When dealing with a whole number, a fraction, or a decimal, add the additive inverse to both sides of the equation to isolate the variable.
As a rule, you can remember additive inverses as having the opposite sign (positive or negative) as the number. This is true with fractions as well.
Also remember, in order to add or subtract fractions, they must have a common denominator.

Practice Problems

$-\dfrac{9}{8}\;+\;{\text{g}}=-\dfrac{3}{8}$ (
Solution

x

Solution: $\dfrac{3}{4}$
Details:
In this example, you are solving for the variable g. The only operation involved between g and other numbers is addition. Solve this the same way you solve equations with integers. Unravel the equation using the order of operations backward.

Step 1: Do the inverse of any addition or subtraction.

On the left-hand side of the equal sign $-\dfrac{9}{8}$ is being added to g. Remove this by adding the inverse to both sides of the equation.

Add $+\dfrac{9}{8}$ to both sides.

The left side of the equation:
$-\dfrac{9}{8}+\dfrac{9}{8}=0$ leaving just the variable g.

The right side of the equation:

Since both of these fractions have the same denominator you can add the numerators.

$-\dfrac{3}{8}+\dfrac{9}{8}=\dfrac{6}{8}$

$This is a picture of the equation \frac{-9}{8}+g=-⅜. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left of the dashed line, with “right-hand side” written on the right. \frac{+9}{8} is written below \frac{-9}{8} and below -⅜. There is a horizontal line beneath that indicating that addition has been performed. Below that g=\frac{6}{8} has been written.$

Normally this would be the end of your work since you isolated the solution for variable g, but in this case, you can still simplify the fraction.

6 and 8 both have 2 as a common factor.

$6=(2)( 3)$

$8=(2)(4)$

Divide out the 2’s leaving just $\dfrac{3}{4}$.

${\text{g}}=\dfrac{6}{8}=\dfrac{\left ( 2 \right )\left ( 3 \right )}{\left ( 2 \right )\left ( 4 \right )}=\dfrac{{\color{Red} \cancel{\left ( 2 \right )} }\left ( 3 \right )}{{\color{Red} \cancel{\left ( 2 \right )}}\left ( 4 \right )}=\dfrac{3}{4}$

The final solution is: ${\text{g}} = \dfrac{3}{4}$.

)
${\text{r}}+\dfrac{1}{4}=-1$ (
Solution

x

Solution: $-\dfrac{5}{4}$

)
${\text{x}}+\dfrac{5}{9}=\dfrac{1}{3}$ (
Solution

x

Solution: $-\dfrac{2}{9}$
Details:
In this example, you need to get the variable x all alone. To do this, add the inverse of $\dfrac{5}{9}$ to both sides of the equation.

Add $-\dfrac{5}{9}$ to both sides.

$This is a picture of the equation x+\frac{5}{9}=\frac{1}{3}. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left of the dashed line, with “right-hand side” written on the right. On the second line x+\frac{5}{9}+ -\frac{5}{9}=\frac{1}{3}+ -\frac{5}{9} is written.$

The left side of the equation:

Positive $\dfrac{5}{9}$ plus negative $\dfrac{5}{9}$ equals 0. This leaves just the variable x by itself.

$This is a picture of the equation x+\frac{5}{9}+ -\frac{5}{9}=\frac{1}{3}+ -\frac{5}{9}. There is a vertical dashed line through the equal sign. “Left-hand side” is written on the left of the dashed line, with “right-hand side” written on the right. On the second line x+ 0=\frac{1}{3}+ -\frac{5}{9} is written. There is a horizontal bracket indicating that \frac{5}{9}+ -\frac{5}{9} is equal to 0.$

The right side of the equation:

In order to add $\dfrac{1}{3}+-\dfrac{5}{9}$ you need to first get common denominators.

9 is a multiple of 3, so the least common multiple is 9.

Multiply $\dfrac{1}{3}$ by 1 in the form of $\dfrac{3}{3}$ to change its denominator to 9.

${\text{X}}=\left (\dfrac{1}{3} \right ){\color{Cyan} \left (\dfrac{3}{3} \right )}{\color{DarkGreen} + {\color{Red} -}\dfrac{5}{9}}$

${\text{X}}=\dfrac{1 \cdot {\color{Cyan} 3}}{3 \cdot {\color{Cyan} 3}}{\color{DarkGreen} + {\color{Red} -}\dfrac{5}{9}}$

${\text{X}}={\color{Cyan} \dfrac{3}{9}} {\color{DarkGreen} + {\color{Red} -}\dfrac{5}{9}}$

Now that the fractions have the same denominator, you can add the numerators.

$3+-5=-2$

${\text{X}}=\dfrac{{\color{Cyan} 3}}{9} + {\color{Red} -}\dfrac{{\color{DarkGreen} 5}}{9}$

${\text{X}}=\dfrac{{\color{Cyan} 3+{\color{Red} -}{\color{DarkGreen} 5}}}{9}$

${\text{X}}=\dfrac{{\color{Red} -}{\color{Orange} 2}}{9}$

The final solution is: ${\text{X}} = -\dfrac{2}{9}$.

)
$-3+{\text{g}}=\dfrac{5}{2}$ (
Solution

x

Solution: $\dfrac{11}{2}$

)
$-4={\text{j}}+\dfrac{4}{3}$ (
Video Solution

x

Solution: $\dfrac{-16}{3}$
Details:

(Solving for a Variable on One Side Using Addition and Subtraction with Fractions #5 (02:14 mins) | Transcript)

) | Transcript)
$\dfrac{7}{6}={\text{A}}-\dfrac{1}{2}$ (
Solution

x

Solution: $\dfrac{5}{3}$

)
$\dfrac{-9}{8}=\dfrac{-3}{4}+{\text{a}}$ (
Video Solution

x

Solution: $\dfrac{-3}{8}$
Details:

(Solving for a Variable on One Side Using Addition and Subtraction with Fractions #7 (02:43 mins) | Transcript)

| Transcript)

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