# Chapter 7 - Section 7.5 - Systems of Inequalities - Exercise Set - Page 865: 113

$\begin{cases}y\leq -2x+6\\y\leq -\frac{1}{2}x+3\\x\geq0\\y\geq0 \end{cases}$

#### Work Step by Step

Step 1. Identify two points on the left line segment as $(0,6),(2,2)$; we can write the line equation as $y-6=\frac{2-6}{2-0}(x-0)$ or $y=-2x+6$; thus the part of the shaded area is $y\leq -2x+6$ Step 2. Identify two points on the right line segment as $(2,2),(6,0)$. We can write the line equation as $y=\frac{2-0}{2-6}(x-6)$ or $y=-\frac{1}{2}x+3$. Thus the part of the shaded area is $y\leq -\frac{1}{2}x+3$ Step 3. The shaded area is in the first quadrant. Thus $x\geq0, y\geq0$ Step 4. Putting together the above results, we have $\begin{cases}y\leq -2x+6\\y\leq -\frac{1}{2}x+3\\x\geq0\\y\geq0 \end{cases}$

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