Question

Solve for $x$ and $y:$ $\frac{15}{x-y}+\frac{22}{x+y}=5,\frac{40}{x-y}+\frac{55}{x+y}=13,x\ne y$ and $x\ne -y.$

• A. $x=-8$ and $y=-3$
• B. $x=8$ and $y=8$
• C. $x=3$ and $y=3$
• D. $x=8$ and $y=3$

Answer 1

The correct option is D $x=8$ and $y=3$

$\frac{15}{(x-y)}+\frac{22}{(x+y)}=5,\frac{40}{x-y}+\frac{55}{x+y}=13$

$15(x+y)+22(x-y)=5(x-y)(x+y)......\left(1\right)$

$40(x+y)+55(x-y)=13(x-y)(x+y)......\left(2\right)$
Divided $1$ and $2$

$\frac{15(x+y)+22(x-y)}{40(x+y)+55(x-y)}=\frac{5}{13}$

$195(x+y)+286(x-y)=200(x+y)+275(x-y)$

$5(x+y)=11(x-y)$

$5x+5y=11x-11y$

$6x=16y$
$3x=8y$
put in (1)

$x=8$

$y=3$