Question

Solve the following systems of equations:
$\frac{3}{x+y}+\frac{2}{x-y}=2$
$\frac{9}{x+y}-\frac{4}{x-y}=1$

Answer 1

Let $x+y=u,x-y=v$

Rewriting the equations :

Equation (1) : $\frac{3}{u}+\frac{2}{v}=2$

Equation (2) : $\frac{9}{u}-\frac{4}{v}=1$

Multiply equation (1) by 3 and subtract the resultant equation from equation (2)

$\Rightarrow \frac{9}{u}+\frac{6}{v}=6$

$\Rightarrow \frac{9}{u}-\frac{4}{v}=1$

$\Rightarrow \frac{10}{v}=5$

$\Rightarrow v=2$

Substituting 'v' in above equations, we get ;

$\Rightarrow u=3$

$\Rightarrow x+y=3$

$\Rightarrow x-y=2$

$\Rightarrow 2x=5$

$\Rightarrow x=\frac{5}{2}$

And $y=\frac{1}{2}$