Question

(3x+3)/(x^2+2x+1)+(x-1)/(x^2-1)
help please..... I think that the denomanators would be (x+1)(x+1) right? and (x-1) (x-1)

Answer 1

\[\dfrac{(3x+3)}{(x^2+2x+1)}+\dfrac{(x-1)}{(x^2-1)}\]
You're right about the denominator of the first fraction, that does factor to \((x+1)(x+1)\). The second one, though, doesn't factor to \((x-1)(x-1)\), though, because that would be \((x^2-2x+1)\). It does factor to \((x+1)(x-1)\), from the difference of squares formula. That means you're left with:
\[\dfrac{(3x+3)}{(x+1)(x+1)}+\dfrac{(x-1)}{(x+1)(x-1)}\]
Can you solve it from there?

Answer 2

would the (x-1) on top cancell out with the (x-1) on the bottom?

Answer 3

It would. The next step after that is to make both denominators the same so that you can add the fractions together.

Answer 4

so then would it be 3x+3/x+1

Answer 5

HINTS:
\[
\frac{(3x+3)}{(x+1)(x-1)}=\frac{3(x+1)}{(x+1)(x+1)}
\]

Answer 6

How did you get there?

Answer 7

@thomas5267 where did you get the 3 on the top part of the other equation?

Answer 8

I factored it out. See this.
http://www.purplemath.com/modules/simpfact.htm

Answer 9

oh okay thanks