Solve the following system of equations:
$\frac{3}{x+y} +\frac{2}{x-y}=2$
$\frac{9}{x+y}-\frac{4}{x-y}=1$

Given:

The given system of equations is:

$\frac{3}{x+y} +\frac{2}{x-y}=2$

$\frac{9}{x+y}-\frac{4}{x-y}=1$

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,

$\frac{3}{x+y} +\frac{2}{x-y}=2$

$3u+2v=2$

$3u+2v-2=0$---(i)

$\frac{9}{x+y}-\frac{4}{x-y}=1$

$9u-4v=1$

$9u=1+4v$

$u=\frac{1+4v}{9}$---(ii)

Substituting $u=\frac{1+4v}{9}$ in equation (i), we get,

$3(\frac{1+4v}{9})+2v-2=0$

$\frac{1+4v}{3}=2-2v$

$1+4v=3(2-2v)$

$1+4v=6-6v$

$6v+4v=6-1$

$10v=5$

$v=\frac{5}{10}$

$v=\frac{1}{2}$

Using $v=\frac{1}{2}$ in equation (i), we get,

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

$3u+1-2=0$

$3u-1=0$

$3u=1$

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

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

$\frac{1}{x+y}=\frac{1}{3}$

$\Rightarrow x+y=3$.....(iii)

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

$\Rightarrow x-y=2$.....(iv)

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

$x+y+x-y=3+2$

$2x=5$

$x=\frac{5}{2}$

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

$\frac{5}{2}-y=2$

$y=\frac{5}{2}-2$

$y=\frac{5-2\times2}{2}$

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

Therefore, the solution of the given system of equations is $x=\frac{5}{2}$ and $y=\frac{1}{2}$.   

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Updated on: 10-Oct-2022

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