Question
Solve :
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
✓
Answer
$\begin{aligned}
\frac{4}{x-y}+\frac{1}{x+y} & =3 ; \frac{2}{x-y}-\frac{3}{x+y}=5 \\
4\left(\frac{1}{x-y}\right)+1\left(\frac{1}{x+y}\right) & =3 \ldots \text { (I) } \\
2\left(\frac{1}{x-y}\right)-3\left(\frac{1}{x+y}\right) & =5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x-y}\right)$ by $a$ and $\left(\frac{1}{x+y}\right)$ by $b$ we get
$\begin{aligned}
& 4 a+b=3 \ldots \text { (III) } \\
& 2 a-3 b=5 \ldots \text { (IV) }
\end{aligned}$
On solving these equations we get, $a=1 b=-1$
But $a=\left(\frac{1}{x-y}\right), b=\left(\frac{1}{x+y}\right)$
$\begin{aligned}
\therefore\left(\frac{1}{x-y}\right) & =1,\left(\frac{1}{x+y}\right)=-1 \\
\therefore \quad x-y & =1 \ldots( V ) \\
x+y & =-1 \ldots( VI )
\end{aligned}$
Solving equation (V) and (VI) we get $x=0, y=-1$
$\therefore$ Solution of the given equations is $(x, y)=(0,-1)$
\frac{4}{x-y}+\frac{1}{x+y} & =3 ; \frac{2}{x-y}-\frac{3}{x+y}=5 \\
4\left(\frac{1}{x-y}\right)+1\left(\frac{1}{x+y}\right) & =3 \ldots \text { (I) } \\
2\left(\frac{1}{x-y}\right)-3\left(\frac{1}{x+y}\right) & =5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x-y}\right)$ by $a$ and $\left(\frac{1}{x+y}\right)$ by $b$ we get
$\begin{aligned}
& 4 a+b=3 \ldots \text { (III) } \\
& 2 a-3 b=5 \ldots \text { (IV) }
\end{aligned}$
On solving these equations we get, $a=1 b=-1$
But $a=\left(\frac{1}{x-y}\right), b=\left(\frac{1}{x+y}\right)$
$\begin{aligned}
\therefore\left(\frac{1}{x-y}\right) & =1,\left(\frac{1}{x+y}\right)=-1 \\
\therefore \quad x-y & =1 \ldots( V ) \\
x+y & =-1 \ldots( VI )
\end{aligned}$
Solving equation (V) and (VI) we get $x=0, y=-1$
$\therefore$ Solution of the given equations is $(x, y)=(0,-1)$
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