Question

One of the solutions of the pair of equations $\frac{a}{x}-\frac{b}{y}=0\text{and}\frac{a{b}^2}{x}+\frac{a^{2}b}{y}=a^2+b^2\text{is:}$

• A. $x=a$
• B. $y=-b$
• C. $x=-a$
• D. $x=a{b}^2$

Answer 1

The correct option is A $x=a$

We have
$\frac{a}{x}-\frac{b}{y}=0\Rightarrow \frac{a}{x}=\frac{b}{y}\Rightarrow x=\frac{a}{b}y...\left(i\right)$

$and\frac{a{b}^2}{x}+\frac{a^{2}b}{y}=a^2+b^2...\left(ii\right)$

$Puttingx=\frac{a}{b}yinequation\left(ii\right),weget$

$\frac{a{b}^{2}b}{ay}+\frac{a^{2}b}{y}=a^2+b^2\Rightarrow \frac{b}{y}(b^2+a^2)=a^2+b^2$

$\Rightarrow y=bFrom\left(i\right)x=\frac{a}{b}b=a$
$Hencex=aandy=b$