9/x-5 + 3/x = -15/x^2-5x

First thing I do is find the LCD which i think is x(x-5) or x^2-5x

I eventually get the equation down to 9x+3x-15=-15

at this point I want to say x=0, but knowing that I can take this further im kind of tempted to(and im kind of second guessing myself), and very stuck in the process. If x=0 then there would be "no solution", correct?

i factor a 3 out: 3(2x+x-5)=-15

and this is where im stuck.

thanks for any help given

2. Originally Posted by NecroWinter
9/x-5 + 3/x = -15/x^2-5x

First thing I do is find the LCD which i think is x(x-5) or x^2-5x

I eventually get the equation down to 9x+3x-15=-15
why are you making this so difficult?

9x + 3(x-5) = -15

12x - 15 = 15

12x = 30

x = 5/2

3. i was kind of confused on this because x = 5/2 isnt an answer thats available for this, i assumed i had to factor instead.
the only answers available are 0, no solution, -5 and "0,5"

i probably should have mentioned it but this problem more than likely has to do with the domain

4. Originally Posted by NecroWinter
i was kind of confused on this because x = 5/2 isnt an answer thats available for this, i assumed i had to factor instead.
the only answers available are 0, no solution, -5 and "0,5"

i probably should have mentioned it but this problem more than likely has to do with the domain
my mistake ...

9x + 3(x-5) = -15
12x - 15 = -15

12x = 0

x = 0 ... an invalid solution.