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Q12.

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Found in: Page 274

### Pre-algebra

Book edition Common Core Edition
Author(s) Ron Larson, Laurie Boswell, Timothy D. Kanold, Lee Stiff
Pages 183 pages
ISBN 9780547587776

# Solve the equation.$\frac{1}{4}x-\frac{3}{8}=9$

The solution of$\frac{1}{4}x-\frac{3}{8}=9$ is$x=\frac{75}{4}$ .

See the step by step solution

## Step 1 . Given

The given equation is$\frac{1}{4}x-\frac{3}{8}=9$ .

## Step 2 . To determine

We have to solve the equation$\frac{1}{4}x-\frac{3}{8}=9$.

## Step 3 . Calculation

To solve the equation, we try to get x alone on a side:

$\begin{array}{c}\frac{1}{4}x-\frac{3}{8}=9\\ \frac{1}{4}x=9+\frac{3}{8}\text{}\left[\text{group like terms}\right]\text{}\\ \frac{1}{4}x=\frac{72+3}{8}\text{}\left[\text{write sum of numerator over denominator}\right]\\ \frac{1}{4}x=\frac{75}{8}\text{}\left[\text{add}\right]\\ x=\frac{75}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{multiply both side with 4}\right]\text{}\\ \text{}\\ \text{}\end{array}$

So, the solution of the equation $\frac{1}{4}x-\frac{3}{8}=9$ is$x=\frac{75}{4}$ .