Question

Solve the following simultaneous 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

$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}...\left(Given\right)$
$\frac{1}{2(3x-y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}...\left(Given\right)$
Substituting $\frac{1}{3x+y}=a$ and $\frac{1}{3x-y}=b$ in the above equations,
$\therefore a+b=\frac{3}{4}=4a+4b=3...\left(1\right)$
$\frac{a}{2}-\frac{b}{2}=-\frac{1}{8}=8a-8b=-2...\left(2\right)$
Multiplying equation $\left(1\right)$ by $2$ and and add with eq(2)
$8a+8b=6...\left(3\right)$
$8a-8b=-2$

a = 1/4
Substituting, $a=\frac{1}{4}$ in equation $\left(1\right)$,
$4\times \frac{1}{4}+4b=3$
$1+4b=3$
$4b=3-1$
$4b=2$
$\therefore b=\frac{2}{4}=\frac{1}{2}$
Resubstituting, $a=\frac{1}{4},b=\frac{1}{2}$
$\therefore \frac{1}{3x+y}=a,\frac{1}{3x-y}=b$
$\frac{1}{3x+y}=\frac{1}{4},\frac{1}{3x-y}=\frac{1}{2}$
$\therefore 3x+y=4...\left(4\right)3x-y=2...\left(5\right)$
Solving equation $\left(4\right)$ and equation $\left(5\right)$,
$3x+y=4$
$\underset{–}{3x-y=2}$
$6x=6...\left(Adding\right)$
$x=\frac{6}{6}$
$\therefore x=1$
Substituting $x=1$ in equation $\left(4\right)$,
$3\left(1\right)+y=4$
$y=4-3$
$\therefore y=1$
$\therefore$ Solution Set: $x=1,y=1$.