Question

Solve the following system of equations,

$\frac{1}{2x}-\frac{1}{y}=-1$ and $\frac{1}{x}+\frac{1}{2y}=8$, where $x\ne 0,y\ne 0$.

Answer 1

Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the given equations become

$\frac{u}{2}-v=-1\Rightarrow u-2v=-2$ .(i)

and, $u+\frac{v}{2}=8\Rightarrow 2u+v=16$ .(ii)

Let us eliminate u from equations (i) and (ii). Multiplying equation (i) by $2$, we get
$2u-4v=-4$ ..(iii)
$2u+v=16$ (iv)

Subtracting (iv) from (iii), we get
$-5v=-20\Rightarrow v=4$

Putting $v=4$ in equation (i), we get
$u-8=-2\Rightarrow u=6$

Hence, $x=\frac{1}{u}=\frac{1}{6}$ and $y=\frac{1}{v}=\frac{1}{4}$
So, the solution of the given system of equation is $x=\frac{1}{6},y=\frac{1}{4}$.