Question

$\frac{3}{x+y}+\frac{2}{x-y}=2,\frac{9}{x+y}-\frac{4}{x-y}=1,\mathrm{where}x+y\ne 0\mathrm{and}x-y\ne 0.$

Answer 1

The given equations are:
$\frac{3}{x+y}+\frac{2}{x-y}=2$ ...(i)
$\frac{9}{x+y}-\frac{4}{x-y}=1$ ...(ii)
Putting $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$, we get:
3u + 2v = 2 ...(iii)
9u − 4v = 1 ...(iv)
On multiplying (iii) by 2, we get:
6u + 4v = 4 ...(v)
On adding (iv) and (v), we get:
15u = 5
$\Rightarrow u=\frac{5}{15}=\frac{1}{3}$
$\Rightarrow \frac{1}{x+y}=\frac{1}{3}\Rightarrow x+y=3$ ...(vi)
On substituting $u=\frac{1}{3}$ in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
$\Rightarrow v=\frac{1}{2}$
$\Rightarrow \frac{1}{x-y}=\frac{1}{2}\Rightarrow x-y=2$ ...(vii)
On adding (vi) and (vii), we get:
2x = 5
⇒ $x=\frac{5}{2}$
On substituting $x=\frac{5}{2}$ in (vi), we get:
$$\begin{gathered} \frac{5}{2}+y=3 \\ \Rightarrow y=\left(3-\frac{5}{2}\right)=\frac{1}{2} \end{gathered}$$
Hence, the required solution is $x=\frac{5}{2}\mathrm{and}\mathrm{y}=\frac{1}{2}$.