Question

Find the following sum:

(Problem in comments...)

Answer 1

\[\frac{ 18 }{ x^2-1 }+\frac{ 9 }{ x+1 }\]

Answer 2

x\(^2\) - 1 = (x+1)(x-1) <--this is your common denominator

do you know the rest?

Answer 3

I'm not quite sure yet. What's the next step?

Answer 4

\[\huge \frac{18}{(x+1)(x-1)} + \frac 9{x+1}\]
since x+1 is present in the 18 fraction, it stays the same. the 9 fraction doesnt have x-1 so multiply it

\[\huge \frac{18 + 9(x-1)}{(x+1)(x-1)}\]
the rest is algebra

Answer 5

\[\frac{ 18+9x-9 }{ (x+1)(x-1) } ?\]
Is that what I'm suppose to do?

Answer 6

@ganeshie8 , help :( I just cant understand math.

Answer 7

@Hero , Can you help me, please?

Answer 8

\[\frac{18 - 9 + 9x}{(x + 1)(x - 1)} = \dfrac{9x + 9}{(x + 1)(x - 1)} \] \(\\= \dfrac{9(x + 1)}{(x + 1)(x - 1)} \\= \dfrac{9}{x - 1}\)

Answer 9

ooooh, ok ! Thank you so much !