Solving a system of equations, why aren't the solutions preserved? I have the equations $$6x^2+8xy+4y^2=3$$
$\qquad$ $\qquad$ $\qquad$$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$$\qquad$and $$2x^2+5xy+3y^2=2$$
This question can be found here, and the answer written by "response" went like this:
Multiply the second by 8 to get: $16x^2+40xy+24y^2=16$
Multiply the first by 5 to get: $30x^2+40xy+20y^2=15$
Subtract the two to get: $14x^2−4y^2=−1$
Later, the guy said to disregard his solution because the solutions to the first two equations do not satisfy the third equation. Why does this happen? Thanks.
 A: Your initial two equations are (rotated) ellipses; their intersection can be up to four points.  Your derived equation is a hyperbola, which has infinitely many solutions, including not only the four but many others.
A: It doesn't happen!  And in fact, you've misquoted the answer by user 'response' in that post.  It says:

my suggestion does not work because a solution to the third eqn need not satisfy the original two eqns.

Necessarily, any solution to the first two equations also satisfies the third.
What they actually meant is this: there can certainly be solutions to the third equation which do not satisfy the first two.  This is because there has been some loss of information - you've canceled out terms!
For instance, in this particular case, $(x,y)=(0,\frac{1}{2})$ solves the third equation, but not the first two.
A: Multiply the first equation by 2 and the second by 3, so the equations are equal. Then you have $12x^2+16xy+8y^2=6x^2+15xy+9y^2$, or $6x^2+xy-9y^2=0$. Now divide through by $x^2$, and define $z=\frac{x}{y}$, yielding $z^2-z-6=0$. Thus $z=3$, or $-2$. Substituting back gives $x=\pm\frac23\sqrt{33}$, or $\pm\frac12\sqrt3$, and the corresponding y values from z.
