Question

Solve the following pair of equations by reducing them to a pair of linear equations:

$\frac{1}{(x-1)}+\frac{1}{(y-2)}=2,\frac{6}{(x-1)}-\frac{2}{(y-2)}=1$

• A. $x=\frac{11}{5},y=\frac{13}{11}$
• B. $x=\frac{13}{5},y=\frac{30}{11}$
• C. $x=\frac{4}{5},y=\frac{20}{11}$
• D. None of these

Answer 1

Answer: (B)

$x=\frac{13}{5},y=\frac{30}{11}$

The given equations are

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

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

Let $\frac{1}{x-1}=u$ and $\frac{1}{y-2}=v$

$u+v=2$ .......(i) and

$6u-2v=1$ ......(ii)

Multiply (i) by $6$

$\Rightarrow 6u+6v=12$ ......(iii)

Subtract (ii) from (iii)

$\Rightarrow 8v=11$

$\Rightarrow v=\frac{11}{8}$

Therefore $\frac{1}{y-2}=\frac{11}{8}$

$\Rightarrow y-2=\frac{8}{11}$

$\Rightarrow y=\frac{30}{11}$

Putting $v=\frac{11}{8}$ in (i), we get

$u+\frac{11}{8}=2$

$\Rightarrow u=\frac{5}{8}$

$\Rightarrow \frac{1}{x-1}=\frac{5}{8}$

$\Rightarrow x-1=\frac{8}{5}$

$\Rightarrow x=\frac{13}{5}$

Therefore, $(x,y)=(\frac{13}{5},\frac{30}{11})$