Question

$\frac{40}{x+y}+\frac{2}{x-y}=5,\frac{25}{x+y}-\frac{3}{x-y}=1,\mathrm{where}x+y\ne 0\mathrm{and}x-y\ne 0.$

Answer 1

The given equations are:
$\frac{40}{x+y}+\frac{2}{x-y}=5$ ...(i)
$\frac{25}{x+y}-\frac{3}{x-y}=1$ ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, we get:
40u + 2v = 5 ...(iii)
25u − 3v = 1 ...(iv)
On multiplying (iii) by 3 and (iv) by 2, we get:
120u + 6v = 15 ...(v)
50u − 6v = 2 ...(vi)
On adding (v) and (vi), we get:
170u = 17
$\Rightarrow u=\frac{17}{170}=\frac{1}{10}$
$\Rightarrow \frac{1}{x+y}=\frac{1}{10}\Rightarrow x+y=10$ ...(vii)
On substituting $u=\frac{1}{10}$ in (vi), we get:
5 − 6v = 2 ⇒ 6v = 3
$\Rightarrow v=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \frac{1}{x-y}=\frac{1}{2}\Rightarrow x-y=2$ ...(viii)
On adding (vii) and (viii), we get:
2x = 12
⇒ x = 6
On substituting x = 6 in (vii), we get:
6 + y = 10
⇒ y = (10 − 6) = 4
Hence, the required solution is x = 6 and y = 4.