Solving for a Variable on One Side Using Multiplication and Division with Fractions

Introduction

In this lesson, you will practice necessary skills for solving for a variable on one side of an equation using multiplication, division, and fractions.

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

Solving for a Variable on One Side: Part 5 - Multiplication and Division with Fractions (10:05 mins) | Transcript

Solving for a Variable on One Side Using Multiplication and Division with Fractions

Review the multiplicative inverses of fractions:

Multiplicative Inverse: Number when multiplied to another number equals 1.

When solving equations where the variable is being multiplied by a fraction, follow the same steps as when it was a whole number and multiply both sides by the multiplicative inverse.

The multiplicative inverse is the reciprocal of the original fraction, but the sign stays the same:

$ {\dfrac {2} {5}}$ → $ {\dfrac {5} {2}}$
$ {-\dfrac {1}{3}}$ → $−3$
$ {\dfrac {7} {8}}$ → $ {\dfrac {8} {7}} $

Example 1
Solve for $x$: $\frac{2}{5}x = 8$

Fractions are the same as division. $\frac{2}{5}x=8$ is the same as $(2\div5)x=8$. In order to solve this equation, isolate $x$. This means $x$ needs to be alone on the left-hand side.

If you want to isolate $x$, you need to remove the $\frac{2}{5}$ by turning it into 1. Do this by multiplying both sides of the equation by the multiplicative inverse of $\frac{2}{5}$, which is $\frac{5}{2}$. Solve this equation using the multiplicative inverse:

\begin{align*}\frac{2}{5}x &=8 &\color{navy}\small\text{Simplify each side}\\\\ \frac{2}{5}x \color{green}\mathbf{(\frac{5}{2})} &= \frac{8}{1} \color{green}\mathbf{(\frac{5}{2})} &\color{navy}\small\text{Multiplicative inverse of $\frac{2}{5}$ is $\frac{5}{2}$}\\\\ x &= \frac{8}{1} \color{green}\mathbf{(\frac{5}{2})} &\color{navy}\small\text{ Left side: $\frac{2}{5}$ and $\frac{5}{2}$ will cancel out}\\\\ x &= \frac{40}{2} &\color{navy}\small\text{Multiply}\\\\x&=20 &\color{navy}\small\text{Simplify}\\\\\end{align*}

Remember, you can find the multiplicative inverse of a fraction by switching the locations of the numbers in the numerator and the denominator. $\frac{2}{5}\cdot \frac{5}{2} = 1$

To check the answer, substitute 20 for $x$ in the original equation of $\frac{2}{5}x=8$.

\begin{align*}
\frac{2}{5}x &= 8 &\color{navy}\small\text{Check the value of x}\\\\
\frac{2}{5}(20) &= 8 &\color{navy}\small\text{Substitute 20 in for x}\\\\
\frac{2}{5}\cdot \frac{20}{1} &= 8 &\color{navy}\small\text{Rewrite 20 as $\frac{20}{1}$}\\\\
\frac{40}{5} &= 8 &\color{navy}\small\text{Multiply}\\\\
8&=8 &\color{navy}\small\text{Simplify}\\\\\\\\
\end{align*}

Since the left side is equal to the right side of the equation ($ 8 = 8$) and makes this statement true, $x$ is equal to 20.

In the previous example, you saw an equation that was being multiplied by a fraction. A fraction inherently has division in it. How would you solve something like Example 2?

Example 2
\begin{align*}\frac{x}{3} &= 6 &\color{navy}\small\text{Solve for the value of x}\\\\\frac{1}{3}x &= 6 &\color{navy}\small\text{Left side: $\frac{x}{3}$ is the same as $\frac{1}{3}x$}\\\\\frac{1}{3}x\color{green}\mathbf{(\frac{3}{1})} &= \frac{6}{1} \color{green}\mathbf{(\frac{3}{1})} &\color{navy}\small\text{Multiplicative inverse of $\frac{1}{3}$ is $\frac{3}{1}$}\\\\ x &= \frac{6}{1} \color{green}\mathbf{(\frac{3}{1})} &\color{navy}\small\text{Left side: $\frac{1}{3}$ and $\frac{3}{1}$ will cancel out}\\\\ x &= \frac{18}{1} &\color{navy}\small\text{Multiply}\\\\x &= 18 &\color{navy}\small\text{Simplify}\\\\\end{align*}

To check the answer, substiute 18 for $x$.
\begin{align*} \frac{x}{3} &= 6 &\color{navy}\small\text{Check the value of x}\\\\ \frac{18}{3} &= 6 &\color{navy}\small\text{Substitute 18 in for x}\\\\ 6 &= 6 &\color{navy}\small\text{Simplify} \end{align*}

This is true. Also, $18\div3=6$, so $x=18$ is the correct solution for this equation.

Things to Remember

Fractions are a type of division.
Multiply both sides of the equation by the multiplicative inverse of a coefficient to isolate the variable.

Practice Problems

$-7{\text{M}}=-\dfrac{7}{4}$ (
Solution

x

Solution: $\dfrac{1}{4}$
Details:
In this example, you want to get the variable M all by itself on one side of the equation. Do this by unraveling the equation using the order of operations backward.

The only operation on the same side of the equation as M is multiplication.

Solve for M by multiplying both sides of the equation by the multiplicative inverse of $−7$.

$(-7)(-\dfrac{1}{7})=1$ so you will multiply both sides by $-\dfrac{1}{7}$.

$The equation -7M=\frac{-7}{4}. There is a vertical dashed line through the equal sign separating the left and right sides of the equation. On the second line is the equation is (\frac{-1}{7})(-7M)=(\frac{-1}{7})(\frac{-7}{4}).$

The left side of the equation:

$-\dfrac{1}{7}$ multiplied to $−7$ is 1 since they are the multiplicative inverses of each other.

The right side of the equation:

Multiply across numerators and denominators to get $\dfrac{7}{28}$.

$The equation (\frac{-1}{7})(-7M)=(\frac{-1}{7})(\frac{-7}{4}). There is a vertical dashed line through the equal sign separating the left and right sides of the equation. On the second line is the equation \frac{(-1)\times(-7)}{7 \times 1}(M)=\frac{(-1)\times(-7)}{7 \times 4}. The third line is the equation \frac{7}{7}(M)=\frac{7}{28}. The fourth line is the equation M=\frac{7}{28}.$

The fraction $\dfrac{7}{28}$ simplifies to $\dfrac{1}{4}$.

${\text{M}}={\color{green} \dfrac{7}{28}=\dfrac{7 \times 1}{7 \times 4}=\dfrac{{\color{green} \cancel{7}}\times 1}{{\color{green} \cancel{7}} \times 4}=\dfrac{1}{4}}$

The final solution is: ${\text{M}} = \dfrac{1}{4}$.

)
$\dfrac{6}{5}{\text{B}}=3$ (
Solution

x

Solution: $\dfrac{5}{2}$

)
$-\dfrac{2}{3}{\text{g}}=-1$ (
Video Solution

x

Solution: $\dfrac{3}{2}$
Details:

| Transcript)
$-\dfrac{2}{7}=-\dfrac{3}{2}{\text{x}}$ (
Solution

x

Solution: $\dfrac{4}{21}$
Details:
In this example, you want to solve for the variable X. It doesn’t matter if it is on the right or left side of the equation. Do the exact same process either way.

In order to isolate X, multiply both sides by the multiplicative inverse of $(-\dfrac{3}{2})$.

$(-\dfrac{3}{2})(-\dfrac{2}{3})= 1$ so you will multiply both sides by $(-\dfrac{2}{3})$.

$The equation \frac{-2}{7}=\frac{-3}{2}x. There is a vertical dashed line through the equal sign separating the left and right sides of the equation. On the second line is the equation (\frac{-2}{3})\frac{-2}{7}=(\frac{-2}{3})\frac{-3}{2}x.$

Remember, it doesn’t matter if the negative is in the numerator, denominator, or just out in front of the fraction. As long as there is only one negative sign, the entire fraction is negative.

Now multiply across numerators and denominators to simplify both sides of the equation.

$The equation (\frac{-2}{3})\frac{-2}{7}=(\frac{-2}{3})\frac{-3}{2}x. There is a vertical dashed line through the equal sign separating the left and right sides of the equation. On the second line is the equation \frac{4}{21}=\frac{6}{6}x. On the third line is the equation \frac{4}{21}=1x.$

In this example, the variable X is on the right side of the equal sign. It doesn’t matter which side it is on as long as it is all by itself.

$\dfrac{4}{21}={\text{X}}$

Therefore: ${\text{X}} = \dfrac{4}{21}$

$\dfrac{4}{21}$ cannot be simplified.

The final solution is: ${\text{X}} = \dfrac{4}{21}$.

)
$4{\text{j}}=\dfrac{3}{2}$ (
Solution

x

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

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$-\dfrac{3}{5}=\dfrac{3}{2}{\text{D}}$ (
Solution

x

Solution: $-\dfrac{2}{5}$

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$-\dfrac{{\text{J}}}{3}=-\dfrac{7}{6}$ (
Video Solution

x

Solution: $\dfrac{7}{2}$
Details:

| Transcript)