Answer
$$2$$
Work Step by Step
Given $$x+y=4, \quad x-y=0, \quad y+3 x=4$$
First, we find the intersection points; we get $x=0,\ x=1,\ x= 2$ and
$$ 4-x\geq 4-3x, \ \ \ 0\leq x\leq 1\\
4-x\geq x, \ \ \ 1\leq x\leq 2$$
Then, the area is given by
\begin{aligned}
\text{Area}&= A_1+A_2\\
&=\int_{0}^{1} [4-x -(4-3x)]dx+ \int_{1}^{2} [(x)-(4-x)]dx\\
&=\int_{0}^{1} [2x]dx+ \int_{1}^{2} [2x-4]dx\\
&=2x^2\bigg|_{0}^{1}+(x^2-4x)\bigg|_{1}^{2}\\
&= 2
\end{aligned}
We use the Mathematica program to plot the region.
