Solve the following system of equations by the method of cross-multiplication:
$x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})$
$x+y=2a^2$

Given:

The given system of equations is:

$x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})$

$x+y=2a^2$

 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,

$x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})$

$x[\frac{(a-b)^2+ab}{a-b}]-y[\frac{(a+b)^2-ab}{a+b}]=0$

$x(\frac{a^2+b^2-2ab+ab}{a-b})-y(\frac{a^2+b^2+2ab-ab}{a+b})=0$

$x(\frac{a^2+b^2-ab}{a-b})-y(\frac{a^2+b^2+ab}{a+b})=0$.....(i)

$x+y-2a^2=0$....(ii)

The solution of a linear pair(standard form) of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by,

$\frac{x}{b_1c_2-b_2c_1}=\frac{-y}{a_1c_2-a_2c_1}=\frac{1}{a_1b_2-a_2b_1}$

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

$a_1=(\frac{a^2+b^2-ab}{a-b}), b_1=-(\frac{a^2+b^2+ab}{a+b}), c_1=0$ and $a_2=1, b_2=1, c_2=-2a^2$

Therefore,

$\frac{x}{(-2 a^{2})[-(\frac{a^{2}+b^{2}+a b}{a+b})]-0 \times 1}=\frac{-y}{(-2 a^{2})(\frac{a^{2}+b^{2}-a b}{a-b})-0 \times 1}=\frac{1}{\frac{a^{2}+b^{2}-a b}{a-b}-[-\frac{(a^{2}+b^{2}+a b)}{a+b}]}$ 

$\frac{x}{2 a^{2}\left(\frac{a^{2}+b^{2}+a b}{a+b}\right)}=\frac{y}{\left(2 a^{2}\right)\left(\frac{a^{2}+b^{2}-a b}{a-b}\right)}=\frac{1}{\frac{a^{2}+b^{2}-a b}{a-b}+\frac{a^{2}+b^{2}+a b}{a+b}}$

$\frac{x}{2 a^{2}\left(\frac{a^{2}+b^{2}+a b}{a+b}\right)}=\frac{y}{\left(2 a^{2}\right)\left(\frac{a^{2}+b^{2}-a b}{a-b}\right)}=\frac{\frac{1}{(a+b)\left(a^{2}+b^{2}-a b\right)+(a-b)\left(a^{2}+b^{2}+a b\right)}}{(a-b)(a+b)}$

$\frac{x}{2 a^{2}\left(\frac{a^{2}+b^{2}+a b}{a+b}\right)}=\frac{y}{2 a^{2}\left(\frac{a^{2}+b^{2}-a b}{a-b}\right)}=\frac{1}{\frac{a^{3}+b^{3}+a^{3}-b^{3}}{(a-b)(a+b)}}$

$\frac{x}{2 a^{2}\left(\frac{a^{2}+b^{2}+a b}{a+b}\right)}=\frac{y}{2 a^{2}\left(\frac{a^{2}+b^{2}-a b}{a-b}\right)}=\frac{1}{\frac{2a^3}{(a-b)(a+b)}}$

This implies,

$\frac{x}{2 a^{2}\left(\frac{a^{2}+b^{2}+a b}{a+b}\right)}=\frac{1}{\frac{2a^3}{(a-b)(a+b)}}$

$x =\frac{2 a^{2}\left(a^{2}+b^{2}+a b\right)}{a+b} \times \frac{(a-b)(a+b)}{2 a^{3}}$

$x=\frac{(a-b)\left(a^{2}+b^{2}+a b\right)}{a}$

$x=\frac{a^{3}-b^{3}}{a}$                [$\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$]

$\frac{y}{2 a^{2}\left(\frac{a^{2}+b^{2}-a b}{a-b}\right)}=\frac{1}{\frac{2 a^{3}}{(a-b)(a+b)}}$

$y=\frac{2 a^{2}\left(a^{2}+b^{2}-a b\right)}{a-b} \times \frac{(a-b)(a+b)}{2 a^{3}}$

$y=\frac{(a+b)\left(a^{2}+b^{2}-a b\right)}{a}$

$y=\frac{a^{3}+b^{3}}{a}$               [$\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$]

The solution of the given system of equations is $x=\frac{a^3-b^3}{a}$ and $y=\frac{a^3+b^3}{a}$.

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Updated on: 10-Oct-2022

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