# Looking for a better way to solve this system of equations.....

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.

• Robotic or not, congratulations on solving it! I'm trying to see if there's some obvious simplifying features of the original equations, but nothing's coming to me yet ... Commented May 4, 2018 at 16:09
• I think your method is clever, not robotic. The equations are nearly homogeneous so that substitution stands a good chance of being useful. Commented May 4, 2018 at 16:16
• You can also solve the4 second equation for $x$ $$x=\pm\sqrt{\frac{3y^2}{y-20}}$$ Commented May 4, 2018 at 16:18
• @Dr.SonnhardGraubner How does that help?
– user399078
Commented May 4, 2018 at 16:24
• How does that not help? Commented May 4, 2018 at 16:34

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 $$15x^4(x^2+17^2)(x-19)=0.$$