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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 $y=\frac{1}{3}$.