Question

Solve each of the following systems of equations by the method of cross-multiplication:
a²x+b²y=c²b²x+a²y=a²

Answer 1

$\text{Given equations are}$

$a^2x+b^2y=c^2$

$b^2x+a^2y=c^2$

$\text{It can be written as}$

$a^2x+b^2y-c^2=0$

$b^2x+a^2y-c^2=0$

$\textbf{By cross multiplication rule, we have}$

$\displaystyle\frac{x}{-b^2c^2+a^2c^2}=\frac{y}{-b^2c^2+a^2c^2}=\frac{1}{a^4-b^4}$

$\displaystyle\frac{x}{c^2(a^2-b^2)}=\frac{y}{c^2(a^2-b^2)}=\frac{1}{(a^2)^2-(b^2)^2}$

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

$\displaystyle\frac{x}{c^2}=\frac{y}{c^2}=\frac{1}{a^2+b^2}$

$\displaystyle\,x=\frac{c^2}{a^2+b^2},\;y=\frac{c^2}{a^2+b^2}$

$\therefore\textbf{The solution is}\;\bf\;x=\frac{c^2}{a^2+b^2},\;\;y=\frac{c^2}{a^2+b^2}$

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