Question

Its just a simple inequality question:
12 / (x-3) > (x+1)
But I have some problem not getting the right answer.

Answer 1

\[12>(x-3)(x+1)\]
Then, \[12>x ^{2}-2x-3\]
\[0>x ^{2}-2x-15\]

Answer 2

@Rizki_Zhou , are you sure?

Answer 3

we can cange position,
\[x ^{2}-2x-15<0\]

Answer 4

@saifoo.khan yes, is it false?

Answer 5

@Rizki_Zhou , yea.. idk im getting a feeling.

Answer 6

@saifoo.khan But I think i do it well.

Answer 7

\[\frac{12}{x-3} - x-1 > 0\]
\[\frac{12 +(-x-1)(x-3)}{x-3} > 0\]
\[\frac{-x^2 + 2x + 3 + 12}{x-3} > 0\]
\[\frac{-x^2 + 2x + 15}{x-3} > 0\]

Answer 8

x^2 - 2x - 15 < 0
(x-5)(x+3) <0
x < 5 and x < -3

Therefore,
-3<x<5

Answer 9

Dont forget about the denominator,
\[\frac{1}{x-3} > 0\]\[x - 3 < 0\]\[x < 3\]

Answer 10

Now, the two inequalities are,
-3<x<5, x < 3

Answer 11

okay @saifoo.khan, got it! thank you :)

Answer 12

Welcome. =D