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 $x\ne 1,y\ne 2)$

• A. $x=4$, $y=5$
• B. $x=5$, $y=4$
• C. $x=\frac{1}{3},y=\frac{1}{3}$
• D. $x=8$, $y=5$

Answer 1

The correct option is A

$x=4$, $y=5$

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
$5p+q=2$
$6p-3q=1$
Now, we can solve the pair of equation by method of elimination.
$15p+3q=6$
$6p-3q=1$
On adding both the above equations, we get
$p=\frac{1}{3}$
Now by substituting the value in one of the equation we find $q=\frac{1}{3}$
As we have assumed $p=\frac{1}{x-1}$
Therefore, $p=\frac{1}{x-1}=\frac{1}{3}$
$x=4$
SImilarly we assumed $q=\frac{1}{y-2}$
Hence $\frac{1}{y-2}=\frac{1}{3}$
$y-2=3$
Thus $x=4$ and $y=5$