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

Given:

The given system of equations is:

$\frac{b}{a}x+\frac{a}{b}y=a^2+b^2$

$x+y=2ab$

 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,

$\frac{b}{a}x+\frac{a}{b}y=a^2+b^2$

$\frac{b^2x+a^2y}{ab}=a^2+b^2$

$b^2x+a^2y=ab(a^2+b^2)$

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

$x+y=2ab$

$x+y-2ab=0$........(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,

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

Therefore,

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

$\frac{x}{-2a^3b+a^3b+ab^3}=\frac{-y}{-2ab^3+a^3b+ab^3}=\frac{1}{b^2-a^2}$

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

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

$x=ab$ and $-y=-ab$

$x=ab$ and $y=ab$

The solution of the given system of equations is $x=ab$ and $y=ab$.   

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Updated on: 10-Oct-2022

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