Question

Solve the following systems of equations:
$\frac{6}{x+y}=\frac{7}{x-y}+3$
$\frac{1}{2(x+y)}=\frac{1}{3(x-y)},$
where $x+y\ne 0$ and $x-y\ne 0$

Answer 1

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

$\Rightarrow$ equation (1): $\frac{6}{u}=\frac{7}{v}+3$

$\Rightarrow$equation (2): $\frac{1}{2u}=\frac{1}{3v}$

From Equation (2); $u=\frac{3v}{2}$

Substituting 'u' in equation (1);

$\Rightarrow \frac{6\times 2}{3v}=\frac{7}{v}+3$

$\Rightarrow \frac{-3}{v}=3$

$\Rightarrow v=-1$

And $u=\frac{-3}{2}$

$\Rightarrow x+y=\frac{-3}{2}$

$\Rightarrow x-y=-1$

Adding both equations, we get ;

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

$\Rightarrow x=\frac{-5}{4}$

Substituting x in above equations, we get ;

$\Rightarrow y=\frac{-1}{4}$