# 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 :

$\frac{a x}{b}-\frac{b y}{a}=a+b$

$a x-b y=2 a b$

Solution:

GIVEN:

$\frac{a x}{b}-\frac{b y}{a}=a+b$

$a x-b y=2 a b$

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

Here we have the pair of simultaneous equation

$\frac{a x}{b}-\frac{b y}{a}-(a+b)=0$

$a x-b y-2 a b=0$

By cross multiplication method we get

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

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

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

$\frac{x}{b(b-a)}=\frac{1}{(b-a)}$

$x=b$

For y

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

$\frac{-y}{a(b-a)}=\frac{1}{(b-a)}$

$y=-a$

Hence we get the value of $x=b$ and $y=-a$