# Solve the following systems of equations:

Question:

Solve the following systems of equations:

$\frac{6}{x+y}=\frac{7}{x-y}+3$

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

where $x+y \neq 0$ and $x-y \neq 0$

Solution:

The given equations are:

$\frac{6}{x+y}=\frac{7}{x-y}+3$

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

Let $\frac{1}{x+y}=u$ and $\frac{1}{x-y}=v$ then equations are

$6 u=7 v+3 \ldots(i)$

$\frac{u}{2}=\frac{v}{3}$ ...(ii)

Multiply equation (ii) by 12 and subtract (ii) from (i), we get

Put the value of $v$ in equation $(i)$, we get

$\Rightarrow 6 u=7 \times-1+3$

$\Rightarrow 6 u=-4$

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

Then

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

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

$\frac{1}{x-y}=-1$

$\Rightarrow x-y=-1$

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

Put the value of  in second equation, we get

$6 \times 2+6 y=5 \times 2 y$

$\Rightarrow 12=4 y$

$\Rightarrow y=3$

Hence the value of $x=-\frac{5}{4}$ and $y=-\frac{1}{4}$.
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