Question

Solve the following system of equations in x and y :

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

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

Answer 1

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

$au-bv+0=0$

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

By cross-multiplication, we have,

$\frac{u}{-b\times -(a^2+b^2)-a^{2}b\times 0}=\frac{-v}{a\times -(a^2+b^2)-a{b}^2\times 0}=\frac{1}{a\times a^{2}b-a{b}^2\times -b}$

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

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

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

Now,

$u=\frac{1}{a}=\frac{1}{x}$

$x=a$

And, $v=\frac{1}{b}=\frac{1}{y}$

$y=b$