# Thread: Inequalities problems

1. ## Inequalities problems

Could someone point out what I'm doing wrong here please:

$\displaystyle \frac{3}{x+1} <\frac{2}{x}$

$\displaystyle \frac{3}{x+1} - \frac{2}{x} <0$

$\displaystyle \frac{x-2}{x(x+1)} <0$

From there I know that the function =0 when x=2
So I know that the function is less than 0 when x<2
But this is wrong according to the answers.

Help appreciated!

2. ## Re: Inequalities problems

Originally Posted by elieh
Could someone point out what I'm doing wrong here please:

$\displaystyle \frac{3}{x+1} <\frac{2}{x}$

$\displaystyle \frac{3}{x+1} - \frac{2}{x} <0$

$\displaystyle \frac{x-2}{x(x+1)} <0$

From there I know that the function =0 when x=2
So I know that the function is less than 0 when x<2
But this is wrong according to the answers.

Help appreciated!
Looking at your final fraction, it is less than zero for positive divided by negative
or for negative divided by positive,
so you have a few cases to examine.

The numerator is negative for x < 2,
but the denominator is positive for x and x+1 both positive or both negative.

Similarly, the numerator is positive for x > 2
and the denominator is always positive in this case.

Hence we cannot have positive divided by negative for this fraction.

3. ## Re: Inequalities problems

Completely forgot about the denominator!
Thanks.