Question

$\frac{10}{x+y}+\frac{2}{x-y}=4,\frac{15}{x+y}-\frac{5}{x-y}=2,\mathrm{where}x\ne y,x\ne -y.$

Answer 1

The given equations are:
$\frac{10}{x+y}+\frac{2}{x-y}=4$ ...(i)
$\frac{15}{x+y}-\frac{5}{x-y}=-2$ ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ , we get:
10u + 2v = 4 ...(iii)
15u − 5v = −2 ...(iv)
On multiplying (iii) by 5 and (iv) by 2, we get:
50u + 10v = 20 ...(v)
30u − 10v = −4 ...(vi)
On adding (v) and (vi), we get:
80u = 16
$\Rightarrow u=\frac{16}{80}=\frac{1}{5}$
$\Rightarrow \frac{1}{x+y}=\frac{1}{5}\Rightarrow x+y=5$ ...(vii)
On substituting $u=\frac{1}{5}$ in (iii), we get:
2 + 2v = 4
⇒ 2v = 2
$\Rightarrow v=1$
$\Rightarrow \frac{1}{x-y}=1\Rightarrow x-y=1$ ...(viii)
On adding (vii) and (viii), we get:
$$\begin{gathered} 2x=6 \\ \Rightarrow x=3 \end{gathered}$$
On substituting $x=3$ in (vii), we get:
$3+y=5\Rightarrow y=2$
Hence, the required solution is $x=3\mathrm{and}y=2$.

Note : There is an error in the question, the correct equations are:
$\frac{10}{x+y}+\frac{2}{x-y}=4$ ...(i)
$\frac{15}{x+y}-\frac{5}{x-y}=-2$ ...(ii)