Question

Solve the following pair of equations.

$\frac{5}{x-1}+\frac{1}{y-2}=2$

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

Here, x $\ne$ 1 and y $\ne$ 2

• A. x = 4, y = 5
• B. x = 5, y = 3
• C. $x=\frac{1}{3},y=\frac{1}{3}$
• D. x = 7, y = 11

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, we get the equations as

$5p+q=2$ .......(1)

$6q-3q=1$ .......(2)

Now, we can solve the pair of equations by method of elimination.

On multiplying the first equation by 3 and then adding it to (2), we get

$15p+3q=6$

$6p-3q=1$

_____________

$p=\frac{1}{3}$

Now by substituting the value of p in equation (2) we get
$q=\frac{1}{3}$

Now, $p=\frac{1}{x-1}$

$\Rightarrow \frac{1}{x-1}=\frac{1}{3}$

$\Rightarrow x=4$

Similarly we assumed $q=\frac{1}{y-2}$

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

$\Rightarrow y-2=3$

$\Rightarrow y=5$

$\therefore x=4andy=5$