Ellen Chang

2022-07-09

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?

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.

Ellen Chang