# 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.

• Note that you quoted him wrongly. He said that "a solution to the third eqn need not satisfy the original two eons." Jul 11, 2013 at 4:18
• @CalvinLin Oh thanks now I see the importance of that, especially in light of vadim's answer
– Ovi
Jul 11, 2013 at 4:21
• As a side note: I would advise on cancelling the constant term instead of the $xy$ term. This will allow you to split the resulting second degree equations into two first degree equations. Jul 11, 2013 at 4:22

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.

• Alpha finds four real (messy) solutions. Jul 11, 2013 at 4:21
• OP misquoted the original solution. Jul 11, 2013 at 4:22
• @CalvinLin Thanks for pointing that out; I'll add it to the answer. Jul 11, 2013 at 4:22

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.

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.