Question:
Solve the following systems of equations:
$\frac{3}{x+y}+\frac{2}{x-y}=2$
$\frac{9}{x+y}-\frac{4}{x-y}=1$
Solution:
The given equations are:
$\frac{3}{x+y}+\frac{2}{x-y}=2$
$\frac{9}{x+y}-\frac{4}{x-y}=1$
Let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ then equations are
$3 u+2 v=2 \ldots(i)$
$9 u-4 v=1 \ldots(i i)$
Multiply equation $(i)$ by 2 and add both equations, we get
Put the value of $u$ in equation $(i)$, we get
$3 \times \frac{1}{3}+2 v=2$
$\Rightarrow 2 v=1$
$\Rightarrow v=\frac{1}{2}$
Then
$\frac{1}{x+y}=\frac{1}{3}$
$\Rightarrow x+y=3$
Add both equations, we get
Put the value of $x$ in first equation, we get
$\frac{5}{2}+y=3$
$\Rightarrow y=\frac{1}{2}$
Hence the value of $x=\frac{5}{2}$ and $y=\frac{1}{2}$.