Solve the following system of equations:
$\frac{5}{x+y} -\frac{2}{x-y}=-1$
$\frac{15}{x+y}+\frac{7}{x-y}=10$

Given:

The given system of equations is:

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

$\frac{15}{x+y}+\frac{7}{x-y}=10$

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,

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

$5u-2v=-1$

$5u-2v+1=0$---(i)

$\frac{15}{x+y}+\frac{7}{x-y}=10$

$15u+7v=10$

$15u=10-7v$

$u=\frac{10-7v}{15}$---(ii)

Substituting $u=\frac{10-7v}{15}$ in equation (i), we get,

$5(\frac{10-7v}{15})-2v+1=0$

$\frac{10-7v}{3}=2v-1$

$10-7v=3(2v-1)$

$10-7v=6v-3$

$6v+7v=10+3$

$13v=13$

$v=\frac{13}{13}$

$v=1$

Using $v=1$ in equation (i), we get,

$5u-2(1)+1=0$

$5u-2+1=0$

$5u-1=0$

$5u=1$

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

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

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

$\Rightarrow x+y=5$.....(iii)

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

$\Rightarrow x-y=1$.....(iv)

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

$x+y+x-y=5+1$

$2x=6$

$x=\frac{6}{2}$

$x=3$

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

$3-y=1$

$y=3-1$

$y=2$

Therefore, the solution of the given system of equations is $x=3$ and $y=2$.  

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Updated on: 10-Oct-2022

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