Given:
$-\frac{96}{x^2y}+1+y=0$
$-\frac{96}{xy^2}+2+x=0$
Solve for $x$ and $y$
How should I find $x$ and $y$?
I thought of using the methods I learned in linear algebra but then I noticed that there is a square in the $x$ and $y$ variable so that makes these set of equations not linear. So, linear algebra doesn't apply here (please correct me if I am wrong).
A method I learned in high school is to divide the two equations together and to substitute it into the other. This is the way that my textbook shows too. But I don't know why this operation is valid i.e. does dividing the two equations really preserve the solution to the system? In linear algebra, as far as I know, dividing two equations together is not a valid row operation.
Therefore, I would like to know, what is the theory behind solving this kind of system of equations.
Well, we can start doing this:
$\frac{96}{x^2y} =1+y$
$\frac{96}{xy^2} =2+x$
Multiply the top equation by $x$ and the bottom by $y$, equate the right hand sides, and subtract out $xy$ gives $x = 2y$. Substituting back into the top equation and rearranging gives
$$y^4 + y^3 = y^3(y+1) = 24,$$
which has solution $y=2$. Then, $x=4$.