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

Given:

The given system of equations is:

$\frac{ax}{b}-\frac{by}{a}=a+b$

$ax-by=2ab$

 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,

$\frac{ax}{b}-\frac{by}{a}=a+b$

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

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

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

$ax-by=2ab$

$ax-by-2ab=0$........(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,

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

Therefore,

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

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

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

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

$x=b$ and $-y=a$

$x=b$ and $y=-a$

The solution of the given system of equations is $x=b$ and $y=-a$.   

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Updated on: 10-Oct-2022

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