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.

Note that you can also write your pair of inequalities as a single linear chain of inequalities:
$$0 \leq x-y \leq x+y \leq 1.$$
So all the information you need is contained in the three inequalities of the chain:
$$\begin{cases}0 \leq x-y,\\x-y \leq x+y,\\x+y \leq 1.\end{cases}
\iff \begin{cases}y \leq x,\\0 \leq y,\\y \leq 1-x.\end{cases}$$
The region bounded by these three lines is found to be the interior of the right isosceles triangle with vertices $(0,0),(0,1),(\frac12,\frac12)$.
