# Solve the following system of equations:$\frac{5}{x-1} +\frac{1}{y-2}=2$$\frac{6}{x-1}-\frac{3}{y-2}=1$

Given:

The given system of equations is:

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

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

To do:

We have to solve the given system of equations.

Solution:

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

This implies, the given system of equations can be written as,

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

$5u+v=2$

$5u+v-2=0$---(i)

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

$6u-3v=1$

$6u=1+3v$

$u=\frac{1+3v}{6}$---(ii)

Substituting $u=\frac{1+3v}{6}$ in equation (i), we get,

$5(\frac{1+3v}{6})+v-2=0$

$\frac{5(1+3v)}{6}=2-v$

$5+15v=6(2-v)$

$5+15v=12-6v$

$15v+6v=12-5$

$21v=7$

$v=\frac{7}{21}$

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

Using $v=\frac{1}{3}$ in equation (i), we get,

$5u+(\frac{1}{3})-2=0$

$5u+\frac{1-2\times3}{3}=0$

$5u+\frac{-5}{3}=0$

$5u=\frac{5}{3}$

$u=\frac{5}{5\times3}$

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

Using the values of $u$ and $v$, we get,

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

$\Rightarrow x-1=3$

$\Rightarrow x=3+1$

$\Rightarrow x=4$

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

$\Rightarrow y-2=3$

$\Rightarrow y=3+2$

$\Rightarrow y=5$

Therefore, the solution of the given system of equations is $x=4$ and $y=5$.

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Updated on: 10-Oct-2022

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