# Solve each of the following systems of equations by the method of cross-multiplication :

Question:

Solve each of the following systems of equations by the method of cross-multiplication :

$6(a x+b y)=3 a+2 b$

$6(b x-a y)=3 b-2 a$

Solution:

GIVEN:

$6(a x+b y)=3 a+2 b$

$6(b x-a y)=3 b-2 a$

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation, after rewriting equations

$6 a x+6 b y-(3 a+2 b)=0$

$6 b x-6 a y-(3 b-2 a)=0$

By cross multiplication method we get

$\frac{x}{-(3 b-2 a)(6 b)-((-6 a) \times-(3 a+2 b))}=\frac{-y}{-(3 b-2 a)(6 a)-((6 b) \times-(3 a+2 b))}=\frac{1}{-36\left(a^{2}+b^{2}\right)}$

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

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

$x=\frac{1}{2}$

Consider the following for y

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

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

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

Hence we get the value of $x=\frac{1}{2}$ and $y=\frac{1}{3}$