Question

Can anyone help me with this?
Solve this equation on y:
(2x-3x^2)/2y = 0

Answer 1

But doesn't that mean that when you multiply both sides with 2y (2y*0) that side becomes zero? Is that "legal"? :P

Answer 2

whoops, sorry, then I'm not sure how to do this either... sorry :/

Answer 3

Ah well, thanks anyway :)

Answer 4

I've got another example:
\[-2x-y \div x+2y = 0\]
and this is supposed to come out as y = -2x
but how?

Answer 5

\[\frac{2x-3x^{2}}{2y} = \frac{x(2-3x)}{2y}=0\]


Your only y solution is that y ≠ 0 :P Your x solutions are that x = 0, and x = 2/3

Answer 6

Oh, crap! I meant x, of course!
But how do you get 2/3?

Answer 7

factor out the x in the numerator then the x in the numerator and the x in the denominator cancel out

Answer 8

It's like a normal quadratic, just hiding :) you have x(2-3x) = 0, so x = 0 and (2 - 3x) =0

Answer 9

Great! Thanks!^^

Answer 10

^^

Answer 11

There isn't an x in the denominator ... :/ ?

Answer 12

When you said, "Oh! I mean x" Gschibby you meant that you wanted the solutions for x right? Not that the 2y was supposed to be 2x? Or I am teh crzy?

Answer 13

No no, I meant I wanted the solutions for x ;)

Answer 14

You're absolutely right!^^

Answer 15

^^

Answer 16

I just posted another example, wich is the one I think jigglypuff answered ;)

Answer 17

gotchya

Answer 18

btw, those two solutions aren't like normal quadratic solutions. it's not a really a parabola....