Intermediate Algebra for College Students (7th Edition)

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

Chapter 4 - Section 4.4 - Linear Inequalities in Two Variables - Exercise Set - Page 294: 9

Answer

The graph is shown below.

Work Step by Step

The inequality is $\frac{x}{2}+\frac{y}{3}\lt1$ Step 1:- Replace the inequality symbol with $=$ and graph the linear equation. $\frac{x}{2}+\frac{y}{3}=1$ Plug $y=0$ for the $x−$intercept. $\Rightarrow \frac{x}{2}+\frac{(0)}{3}=1$ Simplify. $\Rightarrow \frac{x}{2}=1$ Multiply both sides by $2$. $\Rightarrow 2\cdot \frac{x}{2}=2\cdot1$ Simplify. $\Rightarrow x=2$ The $x−$intercept is $2$, so the line passes through $A=(2,0)$. Plug $x=0$ for the $y−$intercept. $\Rightarrow \frac{(0)}{2}+\frac{y}{3}=1$ Simplify. $\Rightarrow \frac{y}{3}=1$ Multiply both sides by $3$. $\Rightarrow 3\cdot \frac{y}{3}=3\cdot 1$ Simplify. $\Rightarrow y=3$ The $y−$intercept is $3$, so the line passes through $B=(0,3)$. Draw a straight line through these intercept points. Step 2:- Choose a test point and plug into the inequality. Let the test point $C=(0,0)$. Plug the test point into inequality. $\Rightarrow \frac{(0)}{2}+\frac{(0)}{3}\lt1$ Simplify. $\Rightarrow 0\lt1$ The statement is correct. The solution set contains the test point. Hence, the upper part of the line is the solution set.
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