Question

Find the value of ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }{y}^{\prime }$for the equations $\frac{a^2}{x}-\frac{b^2}{y}=0;\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b$ where $x,y\ne 0.$

• A. $x=a^2,y=b^2$
• B. $x=b^2,y=a^2$
• C. $x=\frac{b}{a},y=\frac{a}{b}$
• D. $x=\frac{1}{b},y=\frac{1}{a}$

Answer 1

The correct option is A $x=a^2,y=b^2$

$\frac{a^2}{x}-\frac{b^2}{y}=0\left(\text{Given}\right)$

$\Rightarrow y{a}^2-x{b}^2=0.....\left(1\right)$

$\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b\left(\text{Given}\right)$

$\Rightarrow a^{2}by+b^{2}ax=(a+b)xy.....\left(2\right)$

Multiplying ${eq}^N\left(1\right)$ by $a$, we get

$a^{3}y-b^{2}ax=0.....\left(3\right)$

Adding ${eq}^n\left(2\right)\&\left(3\right)$, we have

$(a^{2}by+b^{2}ax)+(a^{3}y-b^{2}ax)=(a+b)xy$

$a^{2}by+b^{2}ax+a^{3}y-b^{2}ax=(a+b)xy$

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

$\Rightarrow x=a^2$

Substituting the value of $x$ in ${eq}^n\left(1\right)$, we have

$y{a}^2-a^2{b}^2=0$

$\Rightarrow y=b^2$

Hence the value of $x$ and $y$ are $a^2$ and $b^2$ respectively.

Hence the correct answer is $\left(A\right)x=a^2,y=b^2$.