## Precalculus (10th Edition)

Published by Pearson

# Chapter 1 - Graphs - Chapter Review - Review Exercises - Page 42: 24

#### Answer

The $x-$intercept is $(6,\,0)$. The $y-$intercept is $(0,\,-4)$. The graph of the line is as shown below:

#### Work Step by Step

To find the $x-$intercept, substitute $y=0$ in the equation$2x-3y=12$. $\Rightarrow 2x-3\left( 0 \right)=12$, $\Rightarrow 2x=12$. Divide on both sides by $12$, $\Rightarrow x=6$. The $x-$intercept is $6$, and the point $\left( 6,0 \right)$ is on the graph of the equation. To find the $x-$intercept, substitute $x=0$ in the equation $2x-3y=12$. $\Rightarrow 2\left( 0 \right)-3y=12$, $\Rightarrow -3y=12$. Divide on both sides by $-3$, $\Rightarrow y=-4$. The $y-$intercept is $-4$, and the point $\left( 0,-4 \right)$ is on the graph of the equation. To draw the graph of the equation of the line $2x-3y=12$, plot the points $\left( 0,-4 \right)$ and $\left( 6,0 \right)$, and connect them to each other.

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