Question

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

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

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

Answer 1

The given equations are,

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

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

Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the above system of equations becomes

$au-bv+0=0$

$a{b}^{2}u+a^{2}bv-(a^2+b^2)=0$

By cross-multiplication, we have
$\frac{u}{(-b)\times [-(a^2+b^2\left)\right]-a^{2}b\times 0}=\frac{-v}{a\times [-(a^2+b^2\left)\right]-a{b}^2\times 0}=\frac{1}{a\times a^{2}b-a{b}^2\times (-b)}$

$\Rightarrow \frac{u}{b(a^2+b^2)}=\frac{-v}{-a(a^2+b^2)}=\frac{1}{a^{3}b+a{b}^3}$

$\Rightarrow \frac{u}{b(a^2+b^2)}=\frac{v}{a(a^2+b^2)}=\frac{1}{ab(a^2+b^2)}$

$\Rightarrow \frac{u}{b(a^2+b^2)}=\frac{v}{a(a^2+b^2)}=\frac{1}{ab(a^2+b^2)}$

So, $u=\frac{b(a^2+b^2)}{ab(a^2+b^2)}=\frac{1}{a}$

and, $v=\frac{a(a^2+b^2)}{ab(a^2+b^2)}=\frac{1}{b}$

Now, $u=\frac{1}{a}\Rightarrow \frac{1}{x}=\frac{1}{a}\Rightarrow x=a$

and $v=\frac{1}{b}\Rightarrow \frac{1}{y}=\frac{1}{b}\Rightarrow y=b$

Hence, the solution of the given system of equations is $x=a$ and $y=b$.