Question

Solve for x and y the following system of equations:
$\frac{5}{(x-y)}+\frac{1}{y-2}=2$
$\frac{6}{x-y}-\frac{3}{y-2}=1$

Answer 1

Given pair of linear equations is

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

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

On multiplying Eq.(i) by 3 to make the coefficients equal of second term, we get the equations as

$\frac{15}{x-y}+\frac{2}{(y-2}=6$....(iii)

On substracting Eq.(iii) from Eq.(ii), we get

$\frac{6}{x-1}-\frac{3}{y-2}-\frac{15}{x-1}+\frac{3}{y-2}=6+1$

$\Rightarrow \frac{6}{x+1}-\frac{15}{x+1}=7$

$\Rightarrow \frac{21}{x-1}=7$

$\Rightarrow x-1=3$

$\Rightarrow x=3+1$

$\Rightarrow x=4$

On putting the value of x = 4 in Eq. (Ii)we get

$\frac{6}{4-1}-\frac{3}{y-2}=1$

$\Rightarrow 2-\frac{3}{y-2}=1$

$\Rightarrow -\frac{3}{y-2}=-1$

$\Rightarrow (y-2)=3$

$\Rightarrow y=3+2$

$\Rightarrow y=5$

Hence, $x=4$ and $y=5$ which is the required solution