Solve the following system of equations by the method of cross-multiplication:
$(a+2b)x+(2a-b)y=2$
$(a-2b)x+(2a+b)y=3$


Given:

The given system of equations is:

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

$(a-2b)x+(2a+b)y=3$

 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,

$(a+2b)x+(2a-b)y-2=0$

$(a-2b)x+(2a+b)y-3=0$

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,

$a_1=(a+2b), b_1=(2a-b), c_1=-2$ and $a_2=(a-2b), b_2=(2a+b), c_2=-3$

Therefore,

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

$\frac{x}{-6a+3b+4a+2b}=\frac{-y}{-3a-6b+2a-4b}=\frac{1}{2a^2+ab+4ab+2b^2-2a^2+ab+4ab-2b^2}$

$\frac{x}{-2a+5b}=\frac{-y}{-a-10b}=\frac{1}{10ab}$

$x=\frac{(5b-2a)\times1}{10ab}$ and $-y=\frac{(-a-10b)\times1}{10ab}$

$x=\frac{5b-2a}{10ab}$ and $-y=\frac{-a-10b}{10ab}$

$x=\frac{5b-2a}{10ab}$ and $y=\frac{a+10b}{10ab}$

The solution of the given system of equations is $x=\frac{5b-2a}{10ab}$ and $y=\frac{a+10b}{10ab}$.

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Updated on: 10-Oct-2022

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