Find the area of the region determined by the system: \begin{align} y & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align} 
Find the area of the region determined by the system: \begin{align} y
 & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}

My attempt


*

*Assuming $x>0$


I have the system 
$$\begin{cases}\begin{align}
y &\ge x \\
y & \le -x+3 \\
\end{align}\end{cases}$$


*

*Assuming $x<0$


$$\begin{cases}\begin{align}
y &\ge -x \\
y & \le x+5 \\
\end{align}\end{cases}$$
A relevant interval is $-1<x<0$ because in this interval $y \le -|x+1| +4 $ is still positive while $y \ge |x|$ is negative.
How do I combine now this information to solve the problem ?
I don't see how I can  get something of the form $y \ge a$ and $y \le b$ which give clear bounds about  area... in this problem it seems more complicated.

For $x > 0$, your area is the area enclosed by the graphs of $y = x$ and $y = -x + 3$ between $0$ and their intersection point. Thus, determine $x = -x +3$ which gets you $2x = 3 \Leftrightarrow x = {3\over 2}$ . So the first area is 
$$\int_{0}^{3 \over 2} 2x - 3 dx$$
Process for $x < 0$ analogously.
