Question

Hi here is my question:

Solve: (x-3)/x+3/x2=-6/x^2+2x

For my solutions, I got -2 and 0, but the problem is, when I plug the numbers back in, I have zero in the denominator which I know you can't have. I'm not sure what I did wrong, or is it just undefined?

Answer 1

Your question is not clear without the use of brackets.
Taking it at face value you way you typed it, it is
(x-3)/x + 3/(x^2) = -6/(x^2) + 2x
This will produce a cubic equation.
Judging from the other types of questions you have posted, I concluded that this cannot be the case.

So please retype your equation using brackets so we can tell one term from another.

Answer 2

oops, my mistake

(x-3)/x + 3/(x+2) + 6/(x^2+2x)

Answer 3

oops, my mistake

(x-3)/x + 3/(x+2) + -(6/(x^2+2x))

I missed the negative sign

Answer 4

oops, my mistake

(x-3)/x + 3/(x+2) = -(6/(x^2+2x))

Sorry, missed the =

Answer 5

Ok, I thought so

(x-3)/x + 3/(x+2) = -(6/(x^2+2x)
(x-3)/x + 3/(x+2) = -6/(x(x+2))
multiply each term by x(x+2)

(x-3)(x+2) + 3x = -6
x^2 - x - 6 + 3x + 6 = 0
x^2 + 2x = 0
x(x+2) = 0
x = 0 or x = -2
the same answers you got, good job!

BUT, as you noticed both of these cause the denominator to be zero, thus the term is undefined.
So there is not solution to your equation.

Answer 6

Thanks :D