Solve the following system of equations by the method of cross-multiplication:
$6(ax+by)=3a+2b$
$6(bx-ay)=3b-2a$

Given:

The given system of equations is:

$6(ax+by)=3a+2b$

$6(bx-ay)=3b-2a$

 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,

$6(ax+by)=3a+2b$

$6ax+6by-(3a+2b)=0$......(i)

$6(bx-ay)=3b-2a$

$6bx-6ay-(3b-2a)=0$.......(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,

$a_1=6a, b_1=6b, c_1=-(3a+2b)$ and $a_2=6b, b_2=-6a, c_2=-(3b-2a)$

Therefore,

$\frac{x}{6b\times-(3b-2a)-(-6a)\times-(3a+2b)}=\frac{-y}{6a\times-(3b-2a)-6b\times-(3a+2b)}=\frac{1}{6a\times(-6a)-6b\times (6b)}$

$\frac{x}{-18b^2+12ab-18a^2-12ab}=\frac{-y}{-18ab+12a^2+18ab+12b^2}=\frac{1}{-36a^2-36b^2}$

$\frac{x}{-18(a^2+b^2)}=\frac{-y}{12(a^2+b^2)}=\frac{1}{-36(a^2+b^2)}$

$x=\frac{-18(a^2+b^2)}{-36(a^2+b^2)}$ and $-y=\frac{12(a^2+b^2)}{-36(a^2+b^2)}$

$x=\frac{1}{2}$ and $-y=\frac{-1}{3}$

$x=\frac{1}{2}$ and $y=\frac{1}{3}$

The solution of the given system of equations is $x=\frac{1}{2}$ and $y=\frac{1}{3}$.  

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Updated on: 10-Oct-2022

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