Solve the following system of equations:
$\frac{3}{x}\ –\ \frac{1}{y}\ =\ −9$
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 5$

Given:

The given system of equations is:

$\frac{3}{x}\ –\ \frac{1}{y}\ =\ −9$

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

To do:

We have to solve the given system of equations.

Solution:

Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$

This implies,

The given system of equations can be written as,

$\frac{3}{x}\ -\ \frac{1}{y}\ =\ -9$

$3u-v=-9$-----(i)

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

$2u+3v=5$

$2u=5-3v$

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

Substitute $u=\frac{5-3v}{2}$ in equation (i), we get,

$3(\frac{5-3v}{2})-v=-9$

Multiplying both sides by $2$, we get,

$2(\frac{3(5-3v)}{2})-2(v)=2(-9)$ 

$15-9v-2v=-18$ 

$-11v=-18-15$

$-11v=-33$ 

$v=\frac{-33}{-11}$

$v=3$

This implies,

$u=\frac{5-3(3)}{2}$

$u=\frac{5-9}{2}$

$u=\frac{-4}{2}$

$u=-2$

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

$y=\frac{1}{v}=\frac{1}{3}$ 

Therefore, the solution of the given system of equations is $x=-\frac{1}{2}$ and $y=\frac{1}{3}$. 

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Updated on: 10-Oct-2022

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