Question

Solve

$\frac{30}{x-y}+\frac{44}{x+y}=10$

$\frac{40}{x-y}+\frac{55}{x+y}=13$

• A. x = 8 and y = 3
• B. x = 3 and y = 8
• C. x = 5 and y = 3
• D. x = 3 and y = 5

Answer 1

The correct option is A

x = 8 and y = 3

Given equations:

$\frac{30}{x-y}+\frac{44}{x+y}=10$

$\frac{40}{x-y}+\frac{55}{x+y}=13$

Let $\frac{1}{x-y}$ = a and $\frac{1}{x+y}=b$

Equations reduce to

30a + 44 b = 10 and 40a + 55b = 13

Solving the above two equations we get a = $\frac{1}{5}$ and b = $\frac{1}{11}$

$\Rightarrow \frac{1}{x-y}=\frac{1}{5}\Rightarrow x-y=5$ .....(i)

and $\frac{1}{x+y}=\frac{1}{11}\Rightarrow x+y=11$.....(ii)

Solving (i) and (ii) we get

x = 8 and y = 3