Question

Solve the following system of equation for x and y.
$\frac{5}{x-1}+\frac{1}{y-2}=2,\frac{6}{x-1}-\frac{3}{y-2}=1$

Answer 1

Let $\frac{1}{x-1}=a$ and $\frac{1}{y-2}=b$

$5a+b=2.....\left(1\right)....\times 3$

$6a-3b=1.....\left(2\right)......\times 1$

$15a+3b=6$

$6a-3b$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

$21a=7$

$\therefore a=\frac{1}{3}$

Putting $a=\frac{1}{3}$ in $\left(1\right)$

$\frac{5}{3}+b=2$

or, $b=2-\frac{5}{3}=\frac{1}{3}$

Substituting value of $a$ and $b$

$\frac{1}{x-1}=\frac{1}{3}\Rightarrow x-1=3.....\left(3\right)$

$\frac{1}{y-2}=\frac{1}{3}\Rightarrow y-2=3......\left(4\right)$

From $\left(3\right),x=4$

From $\left(4\right),y=5$