Something stupid that has me confused.

[99.96]

Alright, so you have this equation

This can be written as:


Then:


Now if you sub y=1 and x=1 into the LHS
You get a different answer to that when you sub x=1 and y=1 into the RHS.

[paulsterio]

What do you mean doesn't exist?

[Asx4Life]

What do you mean it doesn't exist? It is a Hyperbola.

[99.96]

I mean if you sub something like x=7 and y=3 into the LHS of

It does not equal to the RHS when you sub x=7 and y=3 into it.

[paulsterio]

It's cause you divided both sides by 12 whilst completing the square

[99.96]

still dont understand.
i divided both sides by 12, but i moved the 1 across, so the equation still equals 0

[Mao]

Doesn't work like that.

Say you have two multivariable, and . only when this is true for all z.

In your case, the two relationships you built your equation from are and . Therefore, these two equations are not equal. In fact, your equation is only true when and .

In your case, when you sub in x and y, you must process this in 3 steps:

Does F(x,y)=0 ?
Does G(x,y)=0 ?
Does F(x,y) = G(x,y) ?

So when you sub in x and y that doesn't give F(x,y)=0, the very premise of your equation fails, because neither left nor right hand side equal to 0.

(e.g. When you sub in (-1.2), you will find LHS=RHS. When you sub in (-1,3), this will not be true)

[99.96]

Ah i get it, cheers

Come to think of it, my logic was pretty stupid because i was basically arguing that if

and
then  :|

[paulsterio]

Ah i get it, cheers

Come to think of it, my logic was pretty stupid because i was basically arguing that if

and
then  :|

Yep, that's your mistake