# Determine the region bounded by the inequalities

Determine the region bounded by the inequalities: $$0 \leq x + y \leq 1 \\ 0 \leq x - y \leq x + y$$

I don't know what to solve for first, so I just added them:

$$0 \leq x \leq 1 + x + y \\$$

I guess I can subtract $x$:

$$-x \leq 0 \leq 1 + y \\$$

Or:

$$-y - 1 \leq 0 \leq x \\$$

So from this inequality, it looks like some area in the 4th quadrant because $x \geq 0$ means everything to the right of the $y$-axis, and $-y - 1 \leq 0$ means $- 1 \leq y$ which is above the line $y = -1$. However, it looks like I'm analyzing incorrectly as the answer says that it is some area above $y = 0$. I'm not sure what I'm doing wrong.

• There are four linear inequalities in two variables. These correspond to half-planes in the plane, and your region is their intersection. Draw each of these half-planes to see what your region looks like. – Servaes May 20 '14 at 18:50

The inequalities are: $$y\le 1-x$$ $$y\ge -x$$ $$y\ge 0$$ $$y\le x$$ You should graph these and determine the region where all the inequalities hold. Here is a picture of what it should look like
$$0 \leq x-y \leq x+y \leq 1.$$
$$\begin{cases}0 \leq x-y,\\x-y \leq x+y,\\x+y \leq 1.\end{cases} \iff \begin{cases}y \leq x,\\0 \leq y,\\y \leq 1-x.\end{cases}$$
The region bounded by these three lines is found to be the interior of the right isosceles triangle with vertices $(0,0),(0,1),(\frac12,\frac12)$.