Question

Solve the following pairs of equations by reducing them to a pair of linear equations.
$\frac{5}{x-1}-\frac{1}{y-2}=2$ & $\frac{6}{x-1}-\frac{3}{y-2}=1$

Answer 1

$\frac{5}{x-1}-\frac{1}{y-2}=2$ .....$\left(1\right)$

$\frac{6}{x-1}-\frac{3}{y-2}=1$ .....$\left(2\right)$

Let $u=\frac{1}{x-1}$ and $v=\frac{1}{y-2}$

then the above equations $\left(1\right)$ and $\left(2\right)$ becomes

$5u-v=2$ .....$\left(3\right)$

$6u-3v=1$ .....$\left(4\right)$

$6\times$ eqn$\left(3\right)-5\times$ eqn$\left(4\right)$ gives

$30u-6v-30u+15v=12-5$

$\Rightarrow 9v=7$

$\Rightarrow v=\frac{9}{7}$ in eqn$\left(3\right)$ we get

Since $v=\frac{1}{y-2}\Rightarrow y-2=\frac{7}{9}$

$\therefore y=\frac{7}{9}+2=\frac{7+18}{9}=\frac{25}{9}$

Substitute $v=\frac{9}{7}$ in $5u-v=2$

$\Rightarrow 5u=2+v=2+\frac{9}{7}=\frac{14+9}{7}=\frac{23}{7}$

$\Rightarrow u=\frac{23}{35}$

$\Rightarrow \frac{1}{x-1}=\frac{23}{35}$

$\Rightarrow x-1=\frac{35}{23}$

$\Rightarrow 23x-23=35$

$\Rightarrow 23x=35+23=58$

$\therefore x=\frac{58}{23}$

Hence $x=\frac{58}{23},y=\frac{25}{9}$