# 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 :

$2 a x+3 b y=a+2 b$

$3 a x+2 b y=2 a+b$

Solution:

GIVEN:

$2 a x+3 b y=a+2 b$

$3 a x+2 b y=2 a+b$

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

Here we have the pair of simultaneous equation

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

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

By cross multiplication method we get

$\frac{x}{(-(2 a+b) \times 3 b)-(2 b \times(-(a+2 b)))}=\frac{-y}{(2 a) \times(-(2 a+b))-((3 a) \times(-(a+2 b)))}=\frac{1}{4 a b-9 a b}$

$\frac{x}{(-(2 a+b) \times 3 b)-(2 b \times(-(a+2 b)))}=\frac{-y}{(2 a) \times(-(2 a+b))-((3 a) \times(-(a+2 b)))}=\frac{1}{-5 a b}$

Now consider

$\frac{x}{(-(2 a+b) \times 3 b)-(2 b \times(-(a+2 b)))}=\frac{1}{-5 a b}$

$-5 a b x=(-(2 a+b) \times 3 b)-(2 b \times(-(a+2 b)))$

$-5 a b x=\left(-6 a b-3 b^{2}\right)-\left(-2 a b-4 b^{2}\right)$

$-5 a b x=-4 a b+b^{2}$

$5 x a b=4 a b-b^{2}$

$\Rightarrow x=\frac{4 a b-b^{2}}{5 a b}$

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

And

$\frac{-y}{(2 a) \times(-(2 a+b))-((3 a) \times(-(a+2 b)))}=\frac{1}{-5 a b}$

$5 y a b=(2 a) \times(-(2 a+b))-((3 a) \times(-(a+2 b)))$

$5 y a b=-4 a^{2}-2 a b-\left(-3 a^{2}-6 a b\right)$

$5 y a b=-a^{2}+4 a b$

$\Rightarrow y=\frac{4 a b-a^{2}}{5 a b}$

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

Hence we get the value of $x=\frac{4 a-b}{5 a}$ and $\mathrm{y}=\frac{4 b-a}{5 b}$