Question

Solve the following pair of equations:
$\frac{5}{x-1}+\frac{1}{y-2}=2$
$\frac{6}{x-1}-\frac{3}{y-2}=1$
(Where $x\ne 1,y\ne 2)$

Answer 1

If we substitute $\frac{1}{x-1}$ as p and $\frac{1}{y-2}$ as q in the given equations. (As $x\ne 1,y\ne 2)$
We get the equations as
$5p+q=2$.....(i)
$6q-3q=1$.....(ii)
Now we can solve the pair of equation by method of elimination.Multiply equation (i) by 3
$15p+3q=6$....(iii)

Adding both (ii) and (iii)
$21p=7$
$p=\frac{1}{3}$
Now by substituting the value of p in (i), we get
$5\times \frac{1}{3}+q=2$
$\Rightarrow q=2-\frac{5}{3}$
$\Rightarrow q=\frac{1}{3}$
As we have assumed
$p=\frac{1}{x-1}$
$\therefore$ $\frac{1}{x-1}=\frac{1}{3}$
$⟹x=4$
Similarly, we assumed
$q=\frac{1}{y-2}$
$\therefore$ $\frac{1}{y-2}=\frac{1}{3}$
$⟹y-2=3ory=5$
Thus, $x=4$, $y=5$