Solve the following system of equations:
$\frac{2}{3x+2y} +\frac{3}{3x-2y}=\frac{17}{5}$
$\frac{5}{3x+2y}+\frac{1}{3x-2y}=2$

Given:

The given system of equations is:

$\frac{2}{3x+2y} +\frac{3}{3x-2y}=\frac{17}{5}$

$\frac{5}{3x+2y}+\frac{1}{3x-2y}=2$

To do:

We have to solve the given system of equations.

Solution:

Let $\frac{1}{3x+2y}=u$ and $\frac{1}{3x-2y}=v$

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

$\frac{2}{3x+2y} +\frac{3}{3x-2y}=\frac{17}{5}$

$2u+3v=\frac{17}{5}$

$5(2u+3v)=17$

$10u+15v=17$

$10u+15v-17=0$---(i)

$\frac{5}{3x+2y}+\frac{1}{3x-2y}=2$

$5u+v=2$

$15(5u+v)=15(2)$    (Multiplying both sides by 15)

$75u+15v=30$

$75u+15v-30=0$---(ii)

Subtracting equation (i) from equation (ii), we get,

$75u+15v-30-(10u+15v-17)=0$

$65u-13=0$

$65u=13$

$u=\frac{13}{65}$

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

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

$10(\frac{1}{5})+15v-17=0$

$2+15v-17=0$

$15v-15=0$

$15v=15$

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

$v=1$

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

$\frac{1}{3x+2y}=\frac{1}{5}$

$\Rightarrow 3x+2y=5$.....(iii)

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

$\Rightarrow 3x-2y=1$.....(iv)

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

$3x+2y+3x-2y=5+1$

$6x=6$

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

$x=1$

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

$3(1)-2y=1$

$2y=3-1$

$2y=2$

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

$y=1$

Therefore, the solution of the given system of equations is $x=1$ and $y=1$.   

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Updated on: 10-Oct-2022

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