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(A) The shaded region is bounded by the lines $x-y=0$ and $x+y=0$. For the line $x-y=0$,testing a point in the shaded region,such as $(1, 0)$,gives $1

Vedclass · Accepted Answer

$x-y \leq 0, x+y \geq 0$

Vedclass · Answer

(A) The shaded region is bounded by the lines $x-y=0$ and $x+y=0$. For the line $x-y=0$,testing a point in the shaded region,such as $(1, 0)$,gives $1-0=1 > 0$. However,the region includes the line itself,so we consider $x-y \leq 0$ for the region below the line $y=x$. Looking at the graph,the shaded region lies below the line $y=x$ (i.e.,$y \geq x$ or $x-y \leq 0$) and above the line $y=-x$ (i.e.,$y \geq -x$ or $x+y \geq 0$). Thus,the inequalities representing the shaded region are $x-y \leq 0$ and $x+y \geq 0$.

The shaded region in the following figure represents the solution set of:

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