Solve system of equations $3|x|+2y=1, 2x-|y|=4$

Question

help with this system of equations \begin{cases} 3|x|+2y=1 \\ 2x-|y|=4 \\ \end{cases} I have no idea.

Answer 1

Setting $x$ from second equation $(x=\frac 1 2 (|y|+4))$ and substituting the first equation, we get $$1.5 \big||y|+4\big|+2y=1.$$

For all $y \in \mathbb{R}$ we get $||y|+4|=|y|+4$ and the equation can be written in the form $1.5 |y|+2y=-5$.

The solution to this equation is $y=-10$.

So the ultimate solution of our system of equations is $(x,y)=(7,-10)$.

Answer 2

Using $a>0\implies |a|=a, a<0\implies |a|=-a$, solve the four systems below and check compatibility:

$$++\begin{cases} 3x+2y=1 \\ 2x-y=4 \\ \end{cases} $$ $$-+\begin{cases} -3x+2y=1 \\ 2x-y=4 \\ \end{cases} $$ $$+-\begin{cases} 3x+2y=1 \\ 2x+y=4 \\ \end{cases} $$ $$--\begin{cases} -3x+2y=1 \\ 2x+y=4 \\ \end{cases} $$