Question

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

$\frac{a}{x}-\frac{b}{y}=0,\frac{a{b}^2}{x}+\frac{a^{2}b}{y}=(a^2+b^2),wherex\ne 0andy\ne 0.$

Answer 1

$\frac{a}{x}-\frac{b}{y}=0$....(1)

$\frac{a{b}^2}{x}+\frac{a^{2}b}{y}-(a^2+b^2)=0$......(2)

Let $\frac{a}{x}=u\frac{b}{y}=v$ put in equ (1)and (2)

then we get

$u-v=0$.......(3)

$b^{2}u+a^{2}v-(a^2+b^2)=0$......(4)

by cross multiplication we get

$\frac{u}{a^2+b^2-a^2\times 0}=\frac{-v}{-(a^2+b^2)-b^2\times 0}=\frac{1}{a^2-(-b^2)}$

$\frac{u}{a^2+b^2}=\frac{-v}{-(a^2+b^2)}=\frac{1}{a^2+b^2}$

$\frac{u|}{a^2+b^2}=\frac{1}{a^2+b^2}$
$u=1$
and
$\frac{-v}{-(a^2+b^2)}=\frac{1}{(a^2+b^2)}$
$v=1$

but
$\frac{a}{x}=u=1$
$x=a$
and
$\frac{y}{b}=v=1$
$y=b$

hence the solution of the given equation is
$x=a$ and $y=b$