Question

Solve the following systems of equations
$\frac{1}{2x}-\frac{1}{y}=-1;\frac{1}{x}+\frac{1}{2y}=8;x\ne 0,y\ne 0$

Answer 1

The given system of equations are
$\frac{1}{2x}-\frac{1}{y}=-1$
and $\frac{1}{x}+\frac{1}{2y}=8$
Let $\frac{1}{x}=a$ and $\frac{1}{y}=b$ then the given equations reduces to
$\frac{1}{2}a-b=-1$ ......(i)
$a+\frac{1}{2}b=8$ ........(ii)
Multiplying equation (i) by $\frac{1}{2}$

$\frac{1}{4}a-\frac{1}{2}b=-\frac{1}{2}$

and then adding with (ii), we get
$\frac{5}{4}a=\frac{15}{2}$
$\Rightarrow a=\frac{15}{2}\times \frac{4}{5}$
$\Rightarrow a=6$
Substituting the value of a in (i), we get
$\frac{1}{2}\times 6-b=-1$
$\Rightarrow 3-b=-1$
$\Rightarrow b=4$
Now $\frac{1}{x}=a$
$\Rightarrow \frac{1}{x}=6$
$\Rightarrow x=\frac{1}{6}$
$\frac{1}{y}=b$
$\Rightarrow \frac{1}{y}=4\Rightarrow y=\frac{1}{4}$