The Question :
Solve for real $x, y$ :
$$xy^2 = 15x^2 + 17xy + 15y^2$$
$$x^2y= 20x^2 + 3y^2$$
My initial attempts involved adding and subtracting the two equations, eliminating one variable, completing some squares and so on. But there was no significant progress as I soon reached a dead end.
The way I was actually able to solve this problem was somewhat unusual : I divided the equations to get $\frac {y}{x} = \frac{15x^2 +17xy + 15y^2}{20x^2 + 3y^2}$. Seemed a very hopeless start, but soon I realised that I could divide the numerator and denominator by $x^2$ on Right Hand Side of the equation, which allowed me to substitute $\frac {y}x= m$. Clearing the denominator, I obtained a cubic equation in $m$ which factored as $(m^2+1)(m-5)=0$. As $x$ and $y$ are real numbers, I rejected the $m^2 +1 = 0$ possibility and got $m=5$ and hence $y = 5x$. After using this relation, I finally got the ordered pair $(x,y) \equiv (19,95)$ as the solution.
I feel that my method is very "robotic" and unnecessarily complicated. Is there any shorter, better, or more elegant way of solving this problem ? I am unable to find any other approach. Thanks in advance.
I think your method is fine. Over the complex numbers we obtain the solutions $$ (x,y)=(0,0),(19,95),(-17i,17),(17i,17), $$ by taking the resultant with respect to $y$, which is $$ 15x^4(x^2+17^2)(x-19)=0. $$