help me!!

- anonymous

help me!!

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- anonymous

with?

- anonymous

solve the inequality (9/x+1)>9

- anonymous

0

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## More answers

- anonymous

thats not one of answer choices

- Hero

ale5, you have not posted the equation correctly. There are two ways to interpret what you have posted.

- Hero

I'll show you what i mean.

- Hero

With the way you have posted it one person could interpret it as:
|dw:1327171939658:dw|
Another person could interpret it as:
|dw:1327171974536:dw|
Do you see the potential confusion here?

- anonymous

oh okay good. I was like it should be that I did it twice

- Hero

cinar, you approach produces no solution

- anonymous

sorry
if question is 9/(x+1)=9 then solution is
-1

- Hero

That isn't the question.

- Hero

x = 1/63 is a solution so the actual solution appears to include fractions close to 0

- anonymous

\[\frac9{x+1}>9\]
\[\frac9{x+1}-9>0\]
\[\frac{9-9x-9}{x+1}>0\]
\[\frac{-9x}{x+1}>0\]
\[\frac x{x+1}<0\]

- anonymous

|dw:1327184116416:dw|

- Hero

I don't believe that is the solution. 1/63 > 0

- anonymous

\[\frac9x+1>9\]
\[\frac9x>8\]
\[\frac{9-8x}{x}>0\]

- anonymous

|dw:1327184323364:dw|

- Mertsj

have you guys noticed that you are a lot more interested in this problem than the asker is? He/she has disappeared.

- Hero

I don't really pay attention to whether or not the original poster is here. I just care if the correct solution is presented.

- anonymous

cinar I already answered it with that but its not it according to the asker. and they probably just guessed.

- anonymous

I solved question either way..

- anonymous

0

- Mertsj

Divide both sides by 9 and get that the reciprocal of x+1 >1. The only way that the reciprocal of a number is greater than 1 is if the number is <1 and >0
So 0

- Mertsj

Another approach is to take the 2 cases
Case 1 x+1>0 implies x>-1 and 9>9x+9 so 0>x Therefore -10 so this yields no solutions and Case 1 yields the only solutions.

- anonymous

x<9/8

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