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

Ellen Chang 2022-07-09 Answered
Solving a system of equations, why aren't the solutions preserved?
6 x 2 + 8 x y + 4 y 2 = 3
and
2 x 2 + 5 x y + 3 y 2 = 2
Multiply the second by 8 to get: 16 x 2 + 40 x y + 24 y 2 = 16
Multiply the first by 5 to get: 30 x 2 + 40 x y + 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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Answers (2)

Jamarcus Shields
Answered 2022-07-10 Author has 17 answers
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 , 1 2 ) solves the third equation, but not the first two.
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Ellen Chang
Answered 2022-07-11 Author has 5 answers
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.
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