Hello /sci/. What's the maximum number of solutions the

I really don't know /sci/ I want beer and pizza! Do science already!

>>8787148
That's how I feel deep down but I know it's hard for you guys, and you have a life outside of work. Something your teachers may not give a shit about.

looks like a system of equations to me, just put it in a matrix and solve it

>>8787155
I've solved it how it's given. I get a positive x, negative x, and y. Three solutions as stated.

I'm told the answer is two solutions, and I think that's bullshit. Just looking for some quick justification.

>>8787166
Always hesitate if you're correct, especially when debating someone else, but believe the mathematics if you check it and it tells you that you are correct.

>>8787142
there is a theorem is Theory of equation according to which there are n no. of solns ( real and complex) for a polynomial having n as the highest power of the x(solution) in the equation....the no. of real powers can be found out using descartes rules of signs

>>8787142
Pose Y = y^3, X = x^2.

Solutions are (x, -y) ; (x, y). Solutions are COUPLES of reals.

>>8787166
Maybe (+x, y) and (-x, y) are considered two solutions?

There are six solutions.

The number of solutions may be only 1, if c=f=0 for example.

>>8787166
What is a solution to that system of equations? Hint, it's not x, -x, or y.

Since the determinant of the matrix
a b
d e
Is not zero, you can multiply both sides by the matrix inverse and get
x^2 = constant
y^3 = other constant
Then it depends on the field you are solving over and whether the right side constants are zero

If neither constant is zero and you are solving over C, you get 6 solutions.

Isn't this bizou's theorem?

>>8788668
If solving over R, and the top constant is positive you get two values for x. You only have one value for y, so the maximum number of solutions is 2 over R