Solve the following system of equations:
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 13$
$\frac{5}{x}\ –\ \frac{4}{y}\ =\ -2$


Given:

The given system of equations is:


$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 13$


$\frac{5}{x}\ –\ \frac{4}{y}\ =\ -2$


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,


$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 13$


$2u+3v=13$-----(i)


$\frac{5}{x}\ -\ \frac{4}{y}\ =\ -2$


$5u-4v=-2$


$5u=4v-2$


$u=\frac{4v-2}{5}$


Substitute $u=\frac{4v-2}{5}$ in equation (i), we get,


$2(\frac{4v-2}{5})+3v=13$

Multiplying both sides by $5$, we get,

$5(\frac{2(4v-2)}{5})+5(3v)=5(13)$ 

$8v-4+15v=65$ 

$23v=65+4$

$23v=69$ 

$v=\frac{69}{23}$

$v=3$

This implies,

$u=\frac{4(3)-2}{5}$

$u=\frac{12-2}{5}$

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

$u=2$

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

$y=\frac{1}{v}=\frac{1}{3}$ 


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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