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: 25

Answer

The solution set graph is shown below.

Work Step by Step

The given system of inequalities is $\left\{\begin{matrix} 2x& -5y&\leq&10\\ 3x& -2y & \gt&6 \end{matrix}\right.$ First we graph both inequalities. For $2x -5y\leq10$ Replace the inequality symbol with $=$ and graph the linear equation. $2x -5y=10$ Plug $y=0$ for the $x−$intercept. $\Rightarrow 2x -5(0)=10$ $\Rightarrow 2x=10$ Divide both sides by $2$. $\Rightarrow \frac{2x}{2}=\frac{10}{2}$ Simplify. $\Rightarrow x=5$ The $x−$intercept is $5$, so the line passes through $A=(5,0)$. Plug $x=0$ for the $y−$intercept. $\Rightarrow 2(0) -5y=10$ Simplify. $\Rightarrow -5y=10$ Divide both sides by $-5$. $\Rightarrow \frac{-5y}{-5}=\frac{10}{-5}$ Simplify. $\Rightarrow y=-2$ The $y−$intercept is $-2$, so the line passes through $B=(0,-2)$. Draw a solid straight line through these intercept points because equality is included. Less than symbol with negative sign of the coefficient of $y$ indicates that the upper part of the line is the solution set. For $3x -2y\gt6$ Replace the inequality symbol by $=$ and graph the linear equation. $3x -2y=6$ Plug $y=0$ for the $x−$intercept. $\Rightarrow 3x -2(0)=6$ $\Rightarrow 3x=6$ Divide both sides by $3$. $\Rightarrow \frac{3x}{3}=\frac{6}{3}$ Simplify. $\Rightarrow x=2$ The $x−$intercept is $2$, so the line passes through $C=(2,0)$. Plug $x=0$ for the $y−$intercept. $\Rightarrow 3(0) -2y=6$ Simplify. $\Rightarrow -2y=6$ Divide both sides by $-2$. $\Rightarrow \frac{-2x}{-2}=\frac{6}{-2}$ Simplify. $\Rightarrow y=-3$ The $y−$intercept is $-3$, so the line passes through $D=(0,-3)$. Draw a dashed straight line through these intercept points because equality is not included. Greater than symbol with negative sign of the coefficient of $y$ indicates that the lower part of the line is the solution set. The solution set of the system of inequalities is graphed as the intersection (the overlap) of the two half-planes. The open dot at point $E=(\frac{10}{11},-\frac{18}{11})$ is not in the solution set because it does not satisfy one of the inequalities. The combined graph is shown below.
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