Solving simultaneous equations with 2 unknowns and modulus

Question

PS: Try answering this question without graphing the equations.

If $|x|+x+y=5$ and $x+|y|-y=10$, what is the value of $x+y$? 1, 2, 3, 4 or 5?

Squaring LHS and RHS didn't seem realistic.

$|x|+x+y=5$

$2|x|+2x+2y=10$

$x+|y|-y=10$

$2|x|+2x+2y=x+|y|-y$

$|y|-2|x|=x+3y$

That's as far as I could go initially.

Answer 1

HINT

Let distinguish the cases

• $x\ge 0, \quad y \ge 0 \implies \begin{cases}|x|+x+y=5\implies x+x+y=5\implies y=-15\\\\x+|y|-y=10\implies x+y-y=10\implies x=10\end{cases}$

• $x\ge 0, \quad y < 0\implies \begin{cases}|x|+x+y=5\implies x+x+y=5\\\\x+|y|-y=10\implies x-y-y=10\end{cases} \implies \begin{cases}2x+y=5\\\\x-2y=10\end{cases}\implies \begin{cases}4x+2y=10\\\\2y=x-10\end{cases}\implies \begin{cases}4x+x-10=10\\\\2y=x-10\end{cases}\implies \begin{cases}x=4\\\\2y=4-10\implies y=-3\end{cases}$

• $x< 0, \quad y \ge 0$

• $x < 0, \quad y < 0\implies \begin{cases}|x|+x+y=5\implies -x+x+y=5\implies y=5\\\\x+|y|-y=10\implies x-y-y=10\implies x=20\end{cases}$

Answer 2

If $y \ge 0$, then the second equation gives $x=10$. Then the first equation gives $y=-15$, a contradiction. Hence we have $y<0$.

If $x \le 0$ the first equation shows $y=5$, hence, by the second eqaution , we get $x=10 >0$, a contradiction. Thus we have $x>0$.

Can you proceed ?