Solve the following system of equations by the method of cross-multiplication:
$2(ax-by)+a+4b=0$
$2(bx+ay)+b-4a=0$

Given:

The given system of equations is:

$2(ax-by)+a+4b=0$

$2(bx+ay)+b-4a=0$

 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,

$2(ax-by)+a+4b=0$

$2ax-2by+(a+4b)=0$....(i)

$2(bx+ay)+b-4a=0$

$2bx+2ay+(b-4a)=0$......(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,

$a_1=2a, b_1=-2b, c_1=(a+4b)$ and $a_2=2b, b_2=2a, c_2=b-4a$

Therefore,

$\frac{x}{-2b\times(b-4a)-2a\times(a+4b)}=\frac{-y}{2a\times(b-4a)-2b\times(a+4b)}=\frac{1}{2a\times(2a)-2b\times (-2b)}$

$\frac{x}{-2b^2+8ab-2a^2-8ab}=\frac{-y}{2ab-8a^2-2ab-8b^2}=\frac{1}{4a^2+4b^2}$

$\frac{x}{-2(a^2+b^2)}=\frac{-y}{-8(a^2+b^2)}=\frac{1}{4(a^2+b^2)}$

$x=\frac{-2(a^2+b^2)}{4(a^2+b^2)}$ and $-y=\frac{-8(a^2+b^2)}{4(a^2+b^2)}$

$x=\frac{-1}{2}$ and $-y=-2$

$x=\frac{-1}{2}$ and $y=2$

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

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Updated on: 10-Oct-2022

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