Question

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

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

$\frac{x}{a^2}+\frac{y}{b^2}=2$

Answer 1

Given equations:-

$\frac{x}{a}+\frac{y}{b}=a+b=>bx+ay-ab(a+b)=0\frac{x}{a^2}+\frac{y}{b^2}=2=>b^{2}x+a^{2}y-2a^2{b}^2=0$

Using cross-multiplication method,

$\frac{x}{\left(a\right)(-2a^2{b}^2)-\left(a^2\right)(-ab(a+b))}=\frac{y}{(-ab(a+b))\left(b^2\right)-(-2a^2{b}^2)\left(b\right)}=\frac{1}{\left(b\right)\left(a^2\right)-\left(b^2\right)\left(a\right)}\frac{x}{-2a^3{b}^2+a^{3}b(a+b)}=\frac{y}{-a{b}^3(a+b)+2a^2{b}^3}=\frac{1}{a^{2}b-{ab}^2}\frac{x}{a^{3}b(a-b)}=\frac{y}{a{b}^3(a-b)}=\frac{1}{ab(a-b)}=>\frac{x}{a^{3}b(a-b)}=\frac{1}{ab(a-b)}or,x=a^2=>\frac{y}{a{b}^3(a-b)}=\frac{1}{ab(a-b)}or,y=b^2\therefore x=a^{2}andy=b^2$