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

Given:

The given system of equations is:

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

$\frac{5}{x}\ –\ \frac{4}{y}\ =\ -2$

 To do: 

Here, we have to solve the given system of equations by the method of cross-multiplication.

Solution:  

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

The given system of equations can be written as,

$2u+3v=13$ and $5u-4v=-2$

$2u+3v-13=0$ and $5u-4v+2=0$

The solution of a linear pair(standard form) of equations $a_1u+b_1v+c_1=0$ and $a_2u+b_2v+c_2=0$ is given by,

$\frac{u}{b_1c_2-b_2c_1}=\frac{-v}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

Comparing the above equations with the standard form of the equations, we get,

$a_1=2, b_1=3, c_1=-13$ and $a_2=5, b_2=-4, c_2=2$

Therefore,

$\frac{u}{3\times2-(-4)\times(-13)}=\frac{-v}{2\times2-5\times(-13)}=\frac{1}{2\times(-4)-5\times3}$

$\frac{u}{6-52}=\frac{-v}{4+65}=\frac{1}{-8-15}$

$\frac{u}{-46}=\frac{-v}{69}=\frac{1}{-23}$

$\frac{u}{-46}=\frac{1}{-23}$ and $\frac{-v}{69}=\frac{1}{-23}$

$u=\frac{-46\times1}{-23}$ and $-v=\frac{69\times1}{-23}$

$u=\frac{-46}{-23}$ and $-v=\frac{69}{-23}$

$u=2$ and $-v=-3$

$u=2$ and $v=3$

This implies,

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

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

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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