Question

Solve the following system of equations for $x$ and $y,$ such that

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

• A. $x=b,y=a$
• B. $x=a,y=b$
• C. $x=-a,y=-b$
• D. $x-b,y=-a$

Answer 1

Answer: (B)

$x=a,y=b$

Take $\frac{1}{x}=p$ and $\frac{1}{y}=q$

The reciprocals of $p$ and $q$ will be the values of $x$ and $y$ respectively.

The given equations are

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

$\Rightarrow ap-bq=0$ ........ $\left(i\right)$ and

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

$\Rightarrow a{b}^{2}p+a^{2}bq=a^2+b^2$ ........ $\left(ii\right)$

$\left(i\right)\times b^2$

$\Rightarrow a{b}^{2}p-b^{3}q=0$ ....... $\left(iii\right)$

$\left(ii\right)-\left(iii\right)$, we get

$\Rightarrow q\times (a^{2}b+b^3)=a^2+b^2$

$\therefore q=\frac{a^2+b^2}{a^{2}b+b^3}$

$\Rightarrow y=\frac{a^{2}b+b^3}{a^2{+b}^2}=\frac{b\left(a^2{+b}^2\right)}{a^2{+b}^2}=b$

$\left(i\right)\times a^2$, we get

$\Rightarrow a^{3}p-a^{2}bq=0$ ....... $\left(iv\right)$

$\left(ii\right)+\left(iv\right)$, we get

$\Rightarrow p\times ({ab}^2+a^3)=a^2+b^2$

$\therefore p=\frac{a^2+b^2}{a^{2}b+b^3}$

$\Rightarrow x=\frac{a^{2}b+b^3}{a^2{+b}^2}=\frac{a\left(a^2{+b}^2\right)}{a^2{+b}^2}=a$

$\therefore (x,y)=(a,b)$