This is a little embarassing, but it seems like I do not understand why the answers are what they are for these inequalities.

1. $\displaystyle 3x^2 - 2x-1 \geq 0 $

First, I factor the left-hand side.

$\displaystyle \therefore (3x+1)(x-1) \geq 0 $

So, $\displaystyle x - 1 \geq 0 $ if we divide both sides by '3x+1'.

Thus, $\displaystyle x \geq 1 $

Here's the part I don't understand. Naturally, I think that if we divide 'x-1' from both sides. We would get

$\displaystyle 3x+1 \geq 0 $

which yields

$\displaystyle 3x \geq -1 \therefore x \geq -\frac{1}{3} $ (*)

But this can not be right. By counterexample, let x = 0; this inequality is false.

It should be $\displaystyle x \leq -\frac{1}{3} $.

I believe the reason for this is since for (x-1), if x is a negative number, then (x-1) < 0. So in order for the inequality to be equal then '3x+1' must also be negative (since a negative times a negative equals a positive) which explains why it should be

$\displaystyle 3x + 1 \leq 0 $ and not $\displaystyle 3x + 1 \geq 0 $

Hence we solve, we should get $\displaystyle x \leq 1/3 $. What am I doing wrong at (*)?

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2. Another confusing inequality.

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

I multiply both sides by '3-x'

$\displaystyle 1 < -2(3-x) $

$\displaystyle 1 < -6+2x $

$\displaystyle 7 < 2x$

$\displaystyle \frac{7}{2} < x $ (*)

This answer makes no sense. If x = 4, then the inequality is false.

Then, I tried a different way. I add both sides by 2

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

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

$\displaystyle \frac{6-2x+1}{3-x} < 0 $

$\displaystyle 7 - 2x < 0 $

So, $\displaystyle x < \frac{7}{2} $. Since, x can not be 3, or else the denominator is zero. I conclude that $\displaystyle 3 < x < \frac{7}{2} $

What am I doing wrong at (*)?

Thank you.