Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 2 - Review Exercises - Page 173: 37

Answer

The graph is shown in the image file.

Work Step by Step

The given equation of the line is $\Rightarrow 4x=8-2y$ Plug $y=0$ for the $x−$intercept. $\Rightarrow 4x=8-2(0)$ Simplify. $\Rightarrow 4x=8$ Divide both sides by $4$. $\Rightarrow \frac{4x}{4}=\frac{8}{4}$ Simplify. $\Rightarrow x=2$ The $x−$intercept is 2, so the line passes through (2,0). Plug x=0 for the y−intercept. $\Rightarrow 4(0)=8-2y$ Simplify. $\Rightarrow 0=8-2y$ Add $2y$ to both sides. $\Rightarrow 0+2y=8-2y+2y$ Simplify. $\Rightarrow 2y=8$ Divide both sides by $2$. $\Rightarrow \frac{2y}{2}=\frac{8}{2}$ Simplify. $\Rightarrow y=4$ The $y−$intercept is $4$, so the line passes through $(0,4)$. Checkpoint plug $y=2$. $\Rightarrow 4x=8-2(2)$ Simplify. $\Rightarrow 4x=8-4$ $\Rightarrow 4x=4$ Divide both sides by $4$. $\Rightarrow \frac{4x}{4}=\frac{4}{4}$ Simplify. $\Rightarrow x=1$ The checkpoint is $(1,2)$. Plot the three points determined above. Draw a straight line through the points $(2,0)$ and $(0,4)$. We notice that the checkpoint $(1,2)$ also belongs to the line, therefore our graph is correct.
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