Question

Solve the following systems of equations by the method of cross-multiplication:
$\frac{a^2}{x}-\frac{b^2}{y}=0$
$\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b,x,y\ne 0$

Answer 1

Given equations:-

$\frac{a^2}{x}-\frac{b^2}{y}=0and\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b$

Let $\frac{1}{x}=pand\frac{1}{y}=q$

$a^{2}p-b^{2}q=0a^{2}bp+b^{2}aq-(a+b)=0$

Using cross-multiplication method,

$\frac{p}{(-b^2)(-(a+b))-\left(b^{2}a\right)\left(0\right)}=\frac{q}{\left(0\right)\left(a^{2}b\right)-(-(a+b))\left(a^2\right)}=\frac{1}{\left(a^2\right)\left(b^{2}a\right)-\left(a^{2}b\right)(-b^2)}\frac{p}{b^2(a+b)}=\frac{q}{a^2(a+b)}=\frac{1}{a^2{b}^2(a+b)}$

$=>\frac{p}{b^2(a+b)}=\frac{1}{a^2{b}^2(a+b)}orp=\frac{1}{a^2}=>\frac{q}{a^2(a+b)}=\frac{1}{a^2{b}^2(a+b)}orq=\frac{1}{b^2}$

$p=\frac{1}{x}=\frac{1}{a^2}=>x=a^{2}q=\frac{1}{y}=\frac{1}{b^2}=>y=b^2$