Question

Solve the following pair of linear equation:
$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4},\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8},3x+y\ne 0,3x-y\ne 0$

Answer 1

$\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}$

$4(3x+y+3x-y)=3\left(\right(3x+y\left)\right(3x-y\left)\right)$

$4(3x+3x)=3(9x^2-y^2)$

$24x=27x^2-3y^2$

$-y^2+9x^2-8x=0$...(1)

$\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8}$

$\frac{1}{(3x+y)}-\frac{1}{(3x-y)}=-\frac{1}{4}$

$4(3x+y-3x-y)=-1\left(\right(3x+y\left)\right(3x-y\left)\right)$

$4(-2y)=-1(9x^2-y^2)$

$-8y=-9x^2+y^2$

$y^2-9x^2+8y=0$...(2)

From (1) and (2), we get

$x=y$

Substituting the value of x in eqn (2), we get

$y^2-9x^2+8y=0$

$y^2-9y^2+8y=0$

$-8y^2+8y=0$

$8y^2=8y$

$\Rightarrow y=1$

$\Rightarrow x=y=1$