Question

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

Answer 1

Given equation are

$\frac{1}{2x}-\frac{1}{y}=-1...\left(i\right)\times \frac{1}{2}$

$\frac{1}{x}+\frac{1}{2y}=8...\left(ii\right)\times ,1x,y\ne 0$

[ $x,y$ both are denominator and symmetric no need to convert into linear equation hence can be eliminated directly]

Multiplying eqn (i) by the coefficient of

$\frac{1}{y}$ in (ii) and vice versa m we have

$\frac{1}{4x}-\frac{1}{2y}=-\frac{1}{2}...\left(iii\right)$

$\frac{1}{x}+\frac{1}{2y}=8....\left(iv\right)$

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$\frac{1}{4x}+\frac{1}{x}=8-\frac{1}{2}$ [ adding eqn $\left(iii\right)$ and $\left(iv\right)$]

$\Rightarrow \frac{1+4}{4x}=\frac{16-1}{2}$

$\Rightarrow 54x=\frac{15}{2}$

$\Rightarrow 15\times 4x=5\times 2$

$\Rightarrow x=\frac{5\times 2}{15\times 4}\Rightarrow x=\frac{1}{6}$

Now , $\frac{1}{x}+\frac{1}{2y}=8$ [ from (iv) ]

$\Rightarrow \frac{1}{\left(\frac{1}{6}\right)}+\frac{1}{2y}=8[\because x=\frac{1}{6}]$

$\Rightarrow 6+\frac{1}{2y}=8\Rightarrow \frac{1}{2y}=8-6$

$\Rightarrow \frac{1}{2y}=\frac{2}{1}\Rightarrow 4y=1\Rightarrow =y=\frac{1}{4}$

So, $x=\frac{1}{6}$ and $y=\frac{1}{4}$.