# Solving a system of equations, why aren't the solutions preserved? 6 x 2 </msup>

Solving a system of equations, why aren't the solutions preserved?
$6{x}^{2}+8xy+4{y}^{2}=3$
and
$2{x}^{2}+5xy+3{y}^{2}=2$
Multiply the second by $8$ to get: $16{x}^{2}+40xy+24{y}^{2}=16$
Multiply the first by $5$ to get: $30{x}^{2}+40xy+20{y}^{2}=15$
Subtract the two to get: $14{x}^{2}-4{y}^{2}=-1$
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?
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Jamarcus Shields
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, $\left(x,y\right)=\left(0,\frac{1}{2}\right)$ solves the third equation, but not the first two.
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Ellen Chang
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.