Solve the following system of equations:
$\frac{44}{x+y} +\frac{30}{x-y}=10$
$\frac{55}{x+y}+\frac{40}{x-y}=13$

Given:

The given system of equations is:

$\frac{44}{x+y} +\frac{30}{x-y}=10$

$\frac{55}{x+y}+\frac{40}{x-y}=13$

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,

$\frac{44}{x+y} +\frac{30}{x-y}=10$

$44u+30v=10$

$44u+30v-10=0$---(i)

$\frac{55}{x+y}+\frac{40}{x-y}=13$

$55u+40v=13$

$55u=13-40v$

$u=\frac{13-40v}{55}$---(ii)

Substituting $u=\frac{13-40v}{55}$ in equation (i), we get,

$44(\frac{13-40v}{55})+30v-10=0$

$\frac{4(13-40v)}{5}=10-30v$

$52-160v=5(10-30v)$

$52-160v=50-150v$

$160v-150v=52-50$

$10v=2$

$v=\frac{2}{10}$

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

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

$44u+30(\frac{1}{5})-10=0$

$44u+6-10=0$

$44u-4=0$

$44u=4$

$u=\frac{4}{44}$

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

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,

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

$\Rightarrow 2x=16$

$\Rightarrow x=8$

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

$8+y=11$

$\Rightarrow y=11-8$

$\Rightarrow y=3$

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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