Solve the following system of equations by the method of cross-multiplication:
$2ax+3by=a+2b$
$3ax+2by=2a+b$

Given:

The given system of equations is:

$2ax+3by=a+2b$

$3ax+2by=2a+b$

 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,

$2ax+3by-(a+2b)=0$

$3ax+2by-(2a+b)=0$

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,

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

Therefore,

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

$\frac{x}{-6ab-3b^2+2ab+4b^2}=\frac{-y}{-4a^2-2ab+3a^2+6ab}=\frac{1}{4ab-9ab}$

$\frac{x}{-4ab+b^2}=\frac{-y}{-a^2+4ab}=\frac{1}{-5ab}$

$\frac{x}{-4ab+b^2}=\frac{1}{-5ab}$ and $\frac{-y}{-a^2+4ab}=\frac{1}{-5ab}$

$x=\frac{(-4ab+b^2)\times1}{-5ab}$ and $-y=\frac{(-a^2+4ab)\times1}{-5ab}$

$x=\frac{-b(4a-b)}{-5ab}$ and $-y=\frac{-a(a-4b}{-5ab}$

$x=\frac{4a-b}{5a}$ and $-y=\frac{a-4b}{5b}$

$x=\frac{4a-b}{5a}$ and $y=\frac{4b-a}{5b}$

The solution of the given system of equations is $x=\frac{4a-b}{5a}$ and $y=\frac{4b-a}{5b}$.

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Updated on: 10-Oct-2022

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