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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 $y=1$.