Question

Solve: $\frac{20}{x+y}+\frac{3}{x-y}=7$ and $\frac{8}{x-y}-\frac{15}{x+y}=5$

• A. $x=4;y=-7$
• B. $x=4;y=-4$
• C. $x=3;y=2$
• D. $x=1;y=6$

Answer 1

The correct option is C $x=3;y=2$

Let $\frac{20}{x+y}+\frac{3}{x-y}=7$ --- (1)
and $\frac{8}{x-y}-\frac{15}{x+y}=5$ --- (2)

Multiplying equation $\left(1\right)$ with $3$ we get, $\frac{60}{x+y}+\frac{9}{x-y}=21$ ----- equation $\left(3\right)$

Multiplying equation $\left(2\right)$ with $4$ we get, $\frac{32}{x-y}-\frac{60}{x+y}=20$ ----- equation $\left(4\right)$

Adding equations $3$ and $4$, we get $\frac{41}{x-y}=41=>x-y=1$ ---- (5)

Substituting $x-y=1$ in the equation $\left(1\right)$, we get $\frac{20}{x+y}+\frac{3}{1}=7=>\frac{20}{x+y}=4$

=>$x+y=5$ --- (6)

Adding equations $5$ and $6$, we get $2x=6=>x=3$ ---- (5)

Substituting $x=3$ in the equation $\left(5\right)$, we get $3-y=1=>y=2$