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?

$6{x}^{2}+8xy+4{y}^{2}=3$

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

Beginner2022-07-10Added 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, $(x,y)=(0,\frac{1}{2})$ solves the third equation, but not the first two.

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.

Ellen Chang

Beginner2022-07-11Added 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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