Question

If $\left|x\right|+x+y=10$ and $x+\left|y\right|-y=12$ then find the value of '$x+y$'.

• A. $\frac{18}{5}$
• B. $\frac{17}{5}$
• C. $\frac{8}{5}$
• D. None of these

Answer 1

Answer: (A)

$\frac{18}{5}$

Assuming $y>0$ leads to $x=12$ from 2nd equation

as: $x+y-y=12$

$x=12$

and then from 1st Equation $|x|+x+y=10$

$2x+y=10$ or $y=14$

which is contradiction as $y>0$

Assuming $y\le 0$ leads to two cases

(i) if $x\le 0$ then from 1st Equation $y=10$ and

that contradictory again

(2) if $x>0$ then $2x+y=\mathrm{10...}\left(1\right),x-2y=12...\left(2\right)$

$\left(1\right)\times 2+\left(2\right)$

$4x+2y=20$

$x-2y=12$

__________

$5x=32$

$x=\frac{32}{5}$

So on subtituting the value of $x$ in equ (2)

$\frac{32}{5}-2y=12$

$2y=\frac{32}{5}-12$

$y=\frac{-14}{5}$

$\therefore x+y=\frac{32}{5}+\frac{-14}{5}$

$\therefore x+y=\frac{18}{5}$