Solve the following system of equations:
$\frac{1}{2(x+2y)} +\frac{5}{3(3x-2y)}=\frac{-3}{2}$
$\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}$

Given:

The given system of equations is:

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

$\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}$

To do:

We have to solve the given system of equations.

Solution:

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

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

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

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

$\frac{3\times u+5v\times2}{6}=\frac{-3\times3}{2\times3}$

$\frac{3u+10v}{6}=\frac{-9}{6}$

$3u+10v=-9$   (Equating numerators on both sides)

$3u+10v+9=0$---(i)

$\frac{5}{4(x+2y)}-\frac{3}{5(3x-2y)}=\frac{61}{60}$

$\frac{5u}{4}-\frac{3v}{5}=\frac{61}{60}$

$\frac{5u\times5-3v\times4}{20}=\frac{61}{60}$

$\frac{25u-12v}{20}=\frac{61}{60}$

$60(25u-12v)=20(61)$

$3(25u-12v)=61$

$75u-36v=61$

$75u=61+36v$

$u=\frac{61+36v}{75}$---(ii)

Substituting $u=\frac{61+36v}{75}$ in equation (i), we get,

$3(\frac{61+36v}{75})+10v+9=0$

$\frac{61+36v}{25}=-10v-9$

$61+36v=25(-10v-9)$

$61+36v=-250v-225$

$36v+250v=-225-61$

$286v=-286$

$v=\frac{-286}{286}$

$v=-1$

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

$3u+10(-1)+9=0$

$3u-10+9=0$

$3u-1=0$

$3u=1$

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

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

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

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

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

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

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

$x+2y+3x-2y=3+(-1)$

$4x=2$

$x=\frac{2}{4}$

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

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

$3(\frac{1}{2})-2y=-1$

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

$2y=\frac{3+1\times2}{2}$

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

$y=\frac{5}{4}$

Therefore, the solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{5}{4}$.  

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Updated on: 10-Oct-2022

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