Question

$\frac{44}{x+y}+\frac{30}{x-y}=10,\frac{55}{x+y}+\frac{40}{x-y}=13.$

Answer 1

The given equations are:
$\frac{44}{x+y}+\frac{30}{x-y}=10$ ...(i)
$\frac{55}{x+y}+\frac{40}{x-y}=13$ ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ , we get:
44u + 30v = 10 ...(iii)
55u + 40v = 13 ...(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
176u + 120v = 40 ...(v)
165u + 120v = 39 ...(vi)
On subtracting (vi) from (v), we get:
11u = 1
$\Rightarrow u=\frac{1}{11}$
$\Rightarrow \frac{1}{x+y}=\frac{1}{11}\Rightarrow x+y=11$ ...(vii)
On substituting $u=\frac{1}{11}$ in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
$\Rightarrow v=\frac{6}{30}=\frac{1}{5}$
$\Rightarrow \frac{1}{x-y}=\frac{1}{5}\Rightarrow x-y=5$ ...(viii)
On adding (vii) and (viii), we get:
2x = 16
⇒ x = 8
On substituting x = 8 in (vii), we get:
8 + y = 11
⇒ y = (11 − 8) = 3
Hence, the required solution is x = 8 and y = 3.