## Thomas' Calculus 13th Edition

Apply $\quad |x|=\left\{\begin{array}{ll} x, & x\geq 0\\ -x, & x \lt 0 \end{array}\right.$ in each of the four quadrants. We obtain an inequality per quadrant $\left[\begin{array}{l|l|ll} & y\geq 0 & & y \lt 0\\ \hline& & & \\ x\geq 0 & x+y\leq 1 & & x-y\leq 1\\ & (Q.I) & & (Q.IV)\\ \hline& & & \\ x \lt 0 & -x+y\leq 1 & & -x-y\leq 1\\ & (Q.II) & & (Q.III) \end{array}\right]$ In quadrant $I$, $y\leq-x+1 \quad$ test point (0,0) satisfies the inequality. Intercepts $(1,0)$ and $(0,1)$. This is the region below the line $y=-x+1$ In quadrant $II$, $y\leq x+1 \quad \quad$ test point (0,0) satisfies the inequality. Intercepts $(-1,0)$ and $(0,1)$. This is the region below the line $y=x+1$, a triangle. In quadrant $III$, $-x-1\leq y \quad \quad$ test point (0,0) satisfies the inequality. Intercepts $(-1,0)$ and $(0,-1)$. This is the region above the line $y=-x-1$, a triangle. In quadrant $III$, $x-1\leq y \quad \quad$ test point (0,0) satisfies the inequality. Intercepts $(1,0)$ and $(0,-1)$. This is the region above the line $y=x-1$, a triangle. The solution region contains all the above triangles. All the borders are included (graphed with a solid line), since the inequalities include the "$=$" sign.