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Thread: Is this correct? inequality

  1. #1
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    Is this correct? inequality

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

    $\displaystyle ? < x < ?$



    And this is what I did:

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


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


    Therefore the answer being:
    $\displaystyle \frac{-1}{3} < x < 3$

    Is it correct?

    Thanks in advance.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Patrick_John View Post
    The solution of the inequality $\displaystyle \mid 2x - 1 \mid < x + 2$ is

    $\displaystyle ? < x < ?$



    And this is what I did:

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


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


    Therefore the answer being:
    $\displaystyle \frac{-1}{3} < x < 3$

    Is it correct?

    Thanks in advance.
    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.
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  3. #3
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    Thanks!
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  4. #4
    MHF Contributor red_dog's Avatar
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    Hi Patrick!

    You must be more rigurous in such problems.
    First of all we need a condition: $\displaystyle x+2>0\Leftrightarrow x>-2\Leftrightarrow x\in (-2,\infty)$, else the inequality has no solution.
    Now:
    If $\displaystyle 2x-1\geq 0\Leftrightarrow x\geq\frac{1}{2}$ (1) the inequality becomes
    $\displaystyle 2x-1<x+2\Leftrightarrow x<3$ (2)
    From (1) and (2) we have $\displaystyle x\in\left.\left[\frac{1}{2},3\right.\right)$ (3)
    If $\displaystyle 2x-1<0\Leftrightarrow x\in\left(-2,\frac{1}{2}\right)$ (4) the inequality becomes
    $\displaystyle -2x+1<x+2\Leftrightarrow x>-\frac{1}{3}$ (5)
    From (4) and (5) we have $\displaystyle x\in\left(-\frac{1}{3},\frac{1}{2}\right)$ (6)
    Now, from (3) and (6) yields $\displaystyle x\in\left(-\frac{1}{3},3\right)$
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    Hi Patrick!

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

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

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

    case 1:

    $\displaystyle 2x - 1 < x + 2$

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

    case 2:

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

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

    $\displaystyle \Rightarrow 3x > -1$

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


    Putting both conditions into one, we get:

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

    or equivalently, as red_dog said, $\displaystyle x \in \left( - \frac {1}{3},3 \right)$
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  6. #6
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    I will show you all another way to work it.
    $\displaystyle \left| {2x - 1} \right| < x + 2$ implies $\displaystyle 0 < x + 2$.
    Square both sides $\displaystyle 4x^2 - 4x + 1 < x^2 + 4x + 4$.
    Now solve:
    $\displaystyle \begin{array}{l}
    3x^2 - 8x - 3 < 0 \\
    \left( {3x + 1} \right)\left( {x - 3} \right) < 0 \\
    x \in \left( {\frac{{ - 1}}{3},3} \right) \\
    \end{array}$.
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  7. #7
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    Thank you everyone, now I completely understand this!
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