# Thread: Is this correct? inequality

1. ## Is this correct? inequality

The solution of the inequality $\mid 2x - 1 \mid < x + 2$ is

$? < x < ?$

And this is what I did:

For positive value
$2x - 1 = x + 2$
$2x - 1 = x + 2$
$2x - x = 2 + 1$
$x = 3$

For negative value
$2x - 1 = -x - 2$
$2x + x = -2 + 1$
$3x = -1$
$x = \frac{-1}{3}$

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

Is it correct?

2. Originally Posted by Patrick_John
The solution of the inequality $\mid 2x - 1 \mid < x + 2$ is

$? < x < ?$

And this is what I did:

For positive value
$2x - 1 = x + 2$
$2x - 1 = x + 2$
$2x - x = 2 + 1$
$x = 3$

For negative value
$2x - 1 = -x - 2$
$2x + x = -2 + 1$
$3x = -1$
$x = \frac{-1}{3}$

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

Is it correct?

yes, your answer is correct, i don't like how you set out the problem though, with equal signs and such. however, from looking at what you did, i can see that you pretty much have the right concept of what to do.

3. Thanks!

4. Hi Patrick!

You must be more rigurous in such problems.
First of all we need a condition: $x+2>0\Leftrightarrow x>-2\Leftrightarrow x\in (-2,\infty)$, else the inequality has no solution.
Now:
If $2x-1\geq 0\Leftrightarrow x\geq\frac{1}{2}$ (1) the inequality becomes
$2x-1 (2)
From (1) and (2) we have $x\in\left.\left[\frac{1}{2},3\right.\right)$ (3)
If $2x-1<0\Leftrightarrow x\in\left(-2,\frac{1}{2}\right)$ (4) the inequality becomes
$-2x+1-\frac{1}{3}$ (5)
From (4) and (5) we have $x\in\left(-\frac{1}{3},\frac{1}{2}\right)$ (6)
Now, from (3) and (6) yields $x\in\left(-\frac{1}{3},3\right)$

5. Originally Posted by red_dog
Hi Patrick!

You must be more rigurous in such problems.
First of all we need a condition: $x+2>0\Leftrightarrow x>-2\Leftrightarrow x\in (-2,\infty)$, else the inequality has no solution.
Now:
If $2x-1\geq 0\Leftrightarrow x\geq\frac{1}{2}$ (1) the inequality becomes
$2x-1 (2)
From (1) and (2) we have $x\in\left.\left[\frac{1}{2},3\right.\right)$ (3)
If $2x-1<0\Leftrightarrow x\in\left(-2,\frac{1}{2}\right)$ (4) the inequality becomes
$-2x+1-\frac{1}{3}$ (5)
From (4) and (5) we have $x\in\left(-\frac{1}{3},\frac{1}{2}\right)$ (6)
Now, from (3) and (6) yields $x\in\left(-\frac{1}{3},3\right)$
I usually just split it into two inequalities, which is kind of similar to what Patrick did:

$|2x - 1| < x + 2$

We have two cases (one where what is in the absolute values is positive, and one where it's negative):

case 1:

$2x - 1 < x + 2$

$\Rightarrow x < 3$ .............. 1st condition

case 2:

$-(2x - 1) < x + 2$ ................. I think Patrick knows why I made the left side negative, so i won't explain

$\Rightarrow 2x - 1 > -x - 2$

$\Rightarrow 3x > -1$

$\Rightarrow x > - \frac {1}{3}$ ................. 2nd condidtion

Putting both conditions into one, we get:

$- \frac {1}{3} < x < 3$ ..................this is the form the question asked the answer to be put in

or equivalently, as red_dog said, $x \in \left( - \frac {1}{3},3 \right)$

6. I will show you all another way to work it.
$\left| {2x - 1} \right| < x + 2$ implies $0 < x + 2$.
Square both sides $4x^2 - 4x + 1 < x^2 + 4x + 4$.
Now solve:
$\begin{array}{l}
3x^2 - 8x - 3 < 0 \\
\left( {3x + 1} \right)\left( {x - 3} \right) < 0 \\
x \in \left( {\frac{{ - 1}}{3},3} \right) \\
\end{array}$
.

7. Thank you everyone, now I completely understand this!