How to solve a system of equations I have this system of equations:
3x²+7y²=55
and
2x²+7xy=60
Is there a method of solving [x,y] without using x²=t, y²=z?
 A: Yes, since $x$ is nonzero (because of the second equation), we can eliminate
$y=(60-2x^2)/(7x)$ and substitute this in the first equation, which gives
$$
25(x + 4)(x + 3)(x - 3)(x - 4)=0.
$$
A: This is a rather ad hoc solution, and not at all pretty. I notice that if I add twice the second equation to the first, I get
$$\left(3x^2+7y^2\right)+2\left(2x^2+7xy\right)=55+2\cdot60\;,$$
or $7x^2+14xy+7y^2=175$, which readily simplifies to $x+y=\pm5$, or $y=\pm5-x$. Substituting those linear equations into either of the original equations gives you an easy quadratic in $x$.
A: Multiply both sides of the first equation by $12$, of the second by $11$, and subtract. The idea is to obtain a homogeneous equation. 
We get $14x^2-77xy+84y^2=0$. Rewrite as  $(2x-3y)(x-4y)=0$. 
Now express one variable, say $x$, in terms of the other, and substitute back in one of the equations.
Remark: The idea always works for systems of the shape $P(x,y)=a$, $Q(x,y)=b$, where $P(x,y)$ and $Q(x,y)$ are homogeneous quadratics in $x$ and $y$. In general after obtaining the homogeneous quadratic, we will need the Quadratic Formula. 
