SOLUTION: I am trying to solve an equation with fractions. This is the problem
2x over x^2+6x+8 = x over x+4 minus 2 over x+2
(i hope the way i wrote it is not confusing. i'm sorry!)
Algebra.Com
Question 404380: I am trying to solve an equation with fractions. This is the problem
2x over x^2+6x+8 = x over x+4 minus 2 over x+2
(i hope the way i wrote it is not confusing. i'm sorry!)
I have gotten this far. 2x=x^2+2x-2x+4
2x=x^2+4
I can't get any further. I know that it has to be factored again but i am confused. It is not the type of equation that you have to set = to 0. I can do any of the other problems like this but i have never had one end up with an x^2 right there and my teacher told us that when we got to that point we would have to factor it again but i'm not sure if i understand the problem. I also wasn't sure if the 2x-2x was 0 or 1x. Please help me with this problem if you don't mind. Thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2x over x^2+6x+8 = x over x+4 minus 2 over x+2
(i hope the way i wrote it is not confusing. i'm sorry!)
I have gotten this far. 2x=x^2+2x-2x+4
2x=x^2+4
---------------------
(2x)/(x^2+6x+8)= x/(x+4) - 2/(x+2)
-----
Multiply thru by x^2+6x+8 to get:
2x = x(x+2) - 2(x+4)
2x = x^2+2x-2x-8
x^2-2x-8 = 0
---
Factor:
x^2-4x+2x-8 = 0
x(x-4)+2(x-4) = 0
(x-4)(x+2) = 0
---
x = 4 or x = -2
--
x cannot be -2 as that would make the denominator (x+2) = zero.
---
So, x = 4 is the only valid answer.
=================
Cheers,
Stan H.
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