Solve the following system of equations:
$\frac{2}{x}\ +\ \frac{3}{y}\ =\ 2$
$\frac{4}{x}\ –\ \frac{9}{y}\ =\ -1$

Given:

The given system of equations is:

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

$\frac{4}{x}\ –\ \frac{9}{y}\ =\ -1$

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,

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

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

$\frac{4}{x}\ -\ \frac{9}{y}\ =\ -1$

$4u-9v=-1$

$4u=9v-1$

$u=\frac{9v-1}{4}$

Substitute $u=\frac{9v-1}{4}$ in equation (i), we get,

$2(\frac{9v-1}{4})+3v=2$

$\frac{9v-1}{2}+3v=2$

Multiplying both sides by $2$, we get,

$2(\frac{9v-1}{2})+2(3v)=2(2)$ 

$9v-1+6v=4$ 

$15v=1+4$

$15v=5$ 

$v=\frac{5}{15}$

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

This implies,

$u=\frac{9(\frac{1}{3})-1}{4}$

$u=\frac{3-1}{4}$

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

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

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

$y=\frac{1}{v}=\frac{1}{\frac{1}{3}}=3$ 

Therefore, the solution of the given system of equations is $x=2$ and $y=3$. 

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Updated on: 10-Oct-2022

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