Question

Solve the following systems of equations:
$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}$
$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}$

Answer 1

Let $\frac{1}{3x+y}=m,\frac{1}{3x-y}=n$

$\Rightarrow m+n=\frac{3}{4}$

$\Rightarrow m-n=-\frac{1}{4}$

Add above two equations ;

$\Rightarrow 2m=\frac{1}{2}$

$\Rightarrow m=\frac{1}{4}$

$\Rightarrow n=\frac{1}{2}$

$\Rightarrow 3x+y=4$

$\Rightarrow 3x-y=2$

$\Rightarrow 6x=6$

$\Rightarrow x=1$

$\Rightarrow y=1$