Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - Chapter Review - Review Exercises - Page 42: 25

Answer

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

Work Step by Step

To find the $x-$intercept, substitute $y=0$ in the equation$\frac{1}{2}x+\frac{1}{3}y=2$. $\Rightarrow \frac{1}{2}x+\frac{1}{3}\left( 0 \right)=2$ $\Rightarrow \frac{1}{2}x=2$ Multiply on both sides by $2$, $\Rightarrow x=4$ The $x-$intercept is $4$, and the point $\left( 4,0 \right)$ is on the graph of the equation. To find the $y-$intercept, substitute $x=0$ in the equation $\frac{1}{2}x+\frac{1}{3}y=2$. $\Rightarrow \frac{1}{2}\left( 0 \right)+\frac{1}{3}y=2$ $\Rightarrow \frac{1}{3}y=2$ Multiply on both sides by $3$, $\Rightarrow y=6$ The $y-$intercept is $6$, and the point $\left( 0,6 \right)$ is on the graph of the equation. To draw the graph, plot the two intercepts and connect them with a line.
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