Question

Question 1 (iv)
Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{5}{x-1}+\frac{1}{y-2}=2$
$\frac{6}{x-1}-\frac{3}{y-2}=1$

Answer 1

$\frac{5}{x-1}+\frac{1}{y-2}=2$
$\frac{6}{x-1}-\frac{3}{y-2}=1$
Substituting $\frac{1}{x-1}=p$ and $\frac{1}{y-2}=q$ in the given equations, we obtain,
5p + q = 2 ... (i)
6p - 3q = 1 ... (ii)
Now, by multiplying equation (i) by 3 we get,
15p + 3q = 6 ... (iii)
Now, adding equation (ii) and (iii),
21p = 7
$\Rightarrow p=\frac{1}{3}$
Substituting this value in equation (ii) we get,
$6\times \frac{1}{3}-3q=1$
$\Rightarrow$ 2-3q = 1
$\Rightarrow$ -3q = 1-2
$\Rightarrow$ -3q = -1
$\Rightarrow q=\frac{1}{3}$
Now, $p=\frac{1}{x-1}=\frac{1}{3}$
$\Rightarrow \frac{1}{x-1}=\frac{1}{3}$
$\Rightarrow 3=x-1$
$\Rightarrow x=4$
Also, $q=\frac{1}{y-2}=\frac{1}{3}$
$\Rightarrow 3=y-2$
$\Rightarrow y=5$
Hence, x = 4 and y = 5 is the solution.