Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.3 Graph Systems of Linear Inequalities - 3.3 Exercises - Skill Practice - Page 171: 3

Answer

Answer $D$

Work Step by Step

Using the graph we will determine the equations of the two lines. The "red" line has the intercepts $(3,0)$ and $(0,3)$. Its slope is: $$m_1=\dfrac{3-0}{0-3}=-1.$$ Determine the equation of the "red" line using the slope $m_1$ and one of the intercepts, let's say $(0,3)$: $$\begin{align*} y-3&=-(x-0)\\ y&=-x+3\\ x+y&=3. \end{align*}$$ The "blue" line has the intercepts $(4,0)$ and $(0,-4)$. Its slope is: $$m_2=\dfrac{-4-0}{0-4}=1.$$ Determine the equation of the "blue" line using the slope $m_2$ and one of the intercepts, let's say $(0,-4)$: $$\begin{align*} y-(-4)&=x-0\\ -x+y&=-4. \end{align*}$$ The solution of the inequality involving the "red" line is the half plane below the line, so the inequality is: $$x+y<3.\tag1$$ The solution of the inequality involving the "blue" line is the half plane above the line, so the inequality is: $$-x+y>-4.\tag2$$ Using inequalities $(1)$ and $(2)$ we find the system of inequalities: $$\begin{cases} x+y<3\\ -x+y>-4. \end{cases}$$ The correct answer is Answer $D$.
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