Question

Add
(3x+9) / (2x+6) + (8x+2) / (x^2+6x+9)

Answer 1

\[\frac{3(x+3)}{2(x+3)}+\frac{2(x+4)}{(x+3)(x+3)}=\frac{3(x+3)}{2(x+3)}*\frac{(x+3)}{(x+3)}+\frac{2(x+4)}{(x+3)^2}\]
=\[\frac{3(x+3)(x+3)}{2(x+3)^2}+\frac{2}{2}*\frac{2(x+4)}{(x+3)^2}=\frac{3(x^2+6x+9)+4(x+4)}{2(x+3)^2}\]

Answer 2

\[=\frac{3x^2+18x+27+4x+16}{2(x+3)^2}=\frac{3x^2+22x+43}{2(x+3)^2}\]

Answer 3

SORRY, I had a typo, this is the real question

(3x+9) / (2x+6) + (8x+12) / (x^2+6x+9)

Answer 4

Wow, lol

Answer 5

You better be sure it is right because I am solving it this time, lol

Answer 6

Yes, that's right now. lol sorry

Answer 7

(3x+17)/(2x+6)

Answer 8

That can't be right, it gives me a form to fill out its,

_x_+______ (all over)
_x_+______

Answer 9

Yes, you were supposed to enter 3x + 17 over 2x + 6