Solve the following system of equations by the method of cross-multiplication:
$\frac{(x\ +\ y)}{xy}\ =\ 2$
$\frac{(x\ –\ y)}{xy}\ =\ 6$

Given:

The given system of equations is:

$\frac{(x\ +\ y)}{xy}\ =\ 2$

$\frac{(x\ –\ y)}{xy}\ =\ 6$

 To do: 

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

Solution:  

The given system of equations can be written as,

$\frac{(x\ +\ y)}{xy}\ =\ 2$

$\frac{x}{xy}+\frac{y}{xy}-2=0$

$\frac{1}{y}+\frac{1}{x}-2=0$----(i)

$\frac{(x\ -\ y)}{xy}\ =\ 6$

$\frac{x}{xy}-\frac{y}{xy}-6=0$

$\frac{1}{y}-\frac{1}{x}-6=0$----(ii)

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

This implies equations (i) and (ii) can be written as,

$u+v-2=0$ and $-u+v-6=0$

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,

$a_1=1, b_1=1, c_1=-2$ and $a_2=-1, b_2=1, c_2=-6$

Therefore,

$\frac{u}{1\times(-6)-1\times(-2)}=\frac{-v}{1\times(-6)-(-1)\times(-2)}=\frac{1}{1\times1-(-1)\times1}$

$\frac{u}{-6+2}=\frac{-v}{-6-2}=\frac{1}{1+1}$

$\frac{u}{-4}=\frac{-v}{-8}=\frac{1}{2}$

$\frac{u}{-4}=\frac{1}{2}$ and $\frac{-v}{-8}=\frac{1}{2}$

$u=\frac{-4\times1}{2}$ and $-v=\frac{-8\times1}{2}$

$u=\frac{-4}{2}$ and $-v=\frac{-8}{2}$

$u=-2$ and $-v=-4$

$u=-2$ and $v=4$

This implies,

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

$y=\frac{1}{v}=\frac{1}{4}$

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

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Updated on: 10-Oct-2022

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