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.

  • $\begingroup$ As you noted, you can't "divide rows" because the matrix representation for systems of equations works only when the system is linear, and your system is nonlinear. $\endgroup$ – Emily Dec 31 '13 at 20:22
  • $\begingroup$ According to one of my professors, "when you work with non-linear equations you just have to fiddle with it," typically until you can find a way to isolate one variable, then do substitution. Also, one of the reasons division is not valid for working with linear equations is because dividing two linear equations would give you a non-linear equation. The row operations you learn in linear algebra are used because they preserve the "linear-ness" $\endgroup$ – andraiamatrix Dec 31 '13 at 20:32

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, $x=4$.

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  • $\begingroup$ so, there is no theory for systematically solving these kind of equations like there is for systems of linear equations i.e. elementary row operations? $\endgroup$ – mauna Jan 1 '14 at 19:08
  • $\begingroup$ There's a good chance that there has been study. The general problem would involve a system of two equations that can have products of up to square terms in the two variables. That is, two equations of the form $$x^2y^2 + ax^2y + bxy^2 + cxy + dx + ey + f = 0.$$ The specific problem here has some of these coefficients equal to zero. $\endgroup$ – John Jan 2 '14 at 18:27

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