Question

Solve the following pair of equations:

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

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

Answer 1

Answer: (B)

$x=a$ and $y=b$

Given equations are

$\frac{a}{x}-\frac{b}{y}=0$ ....(1)

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

Multiply equation (1) by $a^2$, we get

$\frac{a^3}{x}-\frac{a^{2}b}{y}=0$ ....(3)

Add equations (2) and (3),

$\frac{1}{x}(a{b}^2+a^3)=a^2+b^2$

$\Rightarrow \frac{1}{x}\left[a\right(a^2+b^2\left)\right]=a^2+b^2$

$\Rightarrow x=a$

Substitute this value in equation (1), we get

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

$\Rightarrow 1-\frac{b}{y}=0$

$\Rightarrow y=b$

Therefore, solution is $x=a,y=b$.