Solve the following system of equations:
$\frac{22}{x+y} +\frac{15}{x-y}=5$
$\frac{55}{x+y}+\frac{45}{x-y}=14$

Given:

The given system of equations is:

$\frac{22}{x+y} +\frac{15}{x-y}=5$

$\frac{55}{x+y}+\frac{45}{x-y}=14$

To do:

We have to solve the given system of equations.

Solution:

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

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

$\frac{22}{x+y} +\frac{15}{x-y}=5$

$22u+15v=5$

$22u+15v-5=0$---(i)

$\frac{55}{x+y}+\frac{45}{x-y}=14$

$55u+45v=14$

$55u=14-45v$

$u=\frac{14-45v}{55}$---(ii)

Substituting $u=\frac{14-45v}{55}$ in equation (i), we get,

$22(\frac{14-45v}{55})+15v-5=0$

$\frac{2(14-45v)}{5}=5-15v$

$28-90v=5(5-15v)$

$28-90v=25-75v$

$90v-75v=28-25$

$15v=3$

$v=\frac{3}{15}$

$v=\frac{1}{5}$

Using $v=\frac{1}{5}$ in equation (i), we get,

$22u+15(\frac{1}{5})-5=0$

$22u+3-5=0$

$22u-2=0$

$22u=2$

$u=\frac{2}{22}$

$u=\frac{1}{11}$

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

$\frac{1}{x+y}=\frac{1}{11}$

$\Rightarrow x+y=11$.....(iii)

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

$\Rightarrow x-y=5$.....(iv)

Adding equations (iii) and (iv), we get,

$x+y+x-y=11+5$

$2x=16$

$2x=\frac{16}{2}$

$x=8$

Substituting the value of $x$ in (iv), we get,

$8-y=5$

$y=8-5$

$y=3$

Therefore, the solution of the given system of equations is $x=8$ and $y=3$.   

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Updated on: 10-Oct-2022

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