Question

Solve each of the following systems of equations by the method of cross-multiplication:
(x/a) + (y/b) = a + b
(x/a²) + (y/b²) = 2

Answer 1

$\text{Given equations are}$

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

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

$\text{It can be written as}$

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

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

$\textbf{By multiplication rule}$

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

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

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

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

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

$\implies\displaystyle\;xb^2=ya^2=a^2b^2$

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

$\implies\displaystyle\;x=a^2,\;y=b^2$

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

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