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Thread: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

  1. #1
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    [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Hello all, I think I almost have this but I need to fine tune something...

    Here's the problem:
    $\displaystyle |-3x+4|-4<-1$

    Here's what I did:
    1. $\displaystyle -3<-3x+4<3$

    2. $\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}$

    3. $\displaystyle \frac{7}{3}>x>\frac{1}{3}$

    My questions:
    1. Is my solution correct?
    2. Does the solution need to be rearranged? Like this: $\displaystyle \frac{1}{3}<x<\frac{7}{3}$
    3. What does the solution set look like? Is this acceptable? $\displaystyle [\frac{1}{3},\frac{7}{3}]$


    Thank you.
    -ac
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  2. #2
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    Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Quote Originally Posted by alexcordero View Post
    Hello all, I think I almost have this but I need to fine tune something...

    Here's the problem:
    $\displaystyle |-3x+4|-4<-1$

    Here's what I did:
    1. $\displaystyle -3<-3x+4<3$

    2. $\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}$

    3. $\displaystyle \frac{7}{3}>x>\frac{1}{3}$

    My questions:
    1. Is my solution correct?
    2. Does the solution need to be rearranged? Like this: $\displaystyle \frac{1}{3}<x<\frac{7}{3}$
    3. What does the solution set look like? Is this acceptable? $\displaystyle [\frac{1}{3},\frac{7}{3}]$


    Thank you.
    -ac
    1. Yes it is. Well done!
    2. Yes it is conventional to write your solution as 1/3 < x < 7/3.
    3. You need to use ( ) instead of [ ] to show that 1/3 and 7/3 are excluded.
    Thanks from alexcordero
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  3. #3
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    Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Quote Originally Posted by alexcordero View Post
    Hello all, I think I almost have this but I need to fine tune something...

    Here's the problem:
    $\displaystyle |-3x+4|-4<-1$

    Here's what I did:
    1. $\displaystyle -3<-3x+4<3$

    2. $\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}$

    3. $\displaystyle \frac{7}{3}>x>\frac{1}{3}$

    My questions:
    1. Is my solution correct?
    2. Does the solution need to be rearranged? Like this: $\displaystyle \frac{1}{3}<x<\frac{7}{3}$
    3. What does the solution set look like? Is this acceptable? $\displaystyle [\frac{1}{3},\frac{7}{3}]$
    See Here.
    Thanks from alexcordero
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    Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Quote Originally Posted by alexcordero View Post

    Here's the problem:
    $\displaystyle |-3x+4|-4<-1$

    Here's what I did:
    1. $\displaystyle -3<-3x+4<3$

    2. $\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3} \ \ \ \ $ The directions of the inequality symbols must be reversed. You need extra steps.

    3. $\displaystyle \frac{7}{3}>x>\frac{1}{3}$

    My questions:
    [LIST=1][*]Is my solution correct? No.
    $\displaystyle -3 < -3x + 4 < 3$

    $\displaystyle -7 < -3x < -1 $

    $\displaystyle \frac{-7}{-3} > \frac{-3}{-3}x > \frac{-1}{-3} $

    $\displaystyle \frac{7}{3} > x > \frac{1}{3}$

    $\displaystyle \frac{1}{3} < x < \frac{7}{3}$


    - - - - - - - - - - - - - - - - - OR - - - - - - - - - - - - - - - - -


    $\displaystyle -3 < -3x + 4 < 3$

    $\displaystyle (-1)(-3) > (-1)(-3x + 4) > (-1)(3)$

    $\displaystyle 3 > 3x - 4 > -3 $

    $\displaystyle 7 > 3x > 1 $

    $\displaystyle 1 < 3x < 7 $

    $\displaystyle \frac{1}{3} < x < \frac{7}{3}$



    - - - - - - - - - - - - - - - OR - - -- - - - - - - - - - - - - -


    $\displaystyle -3 < -3x + 4 < 3$

    $\displaystyle -7 < -3x < -1$

    splits into

    $\displaystyle -7 < -3x $

    Add 3x to each side, and add 7 to each side:

    $\displaystyle 3x < 7 $

    $\displaystyle \dfrac{3x}{3} < \dfrac{7}{3}$

    $\displaystyle x < \dfrac{7}{3}$

    and

    $\displaystyle -3x < -1$

    Add 3x to each side, and add 1 to each side:

    $\displaystyle 1 < 3x$

    $\displaystyle \dfrac{1}{3} < \dfrac{3x}{3}$

    $\displaystyle \dfrac{1}{3} < x$

    Putting these together, we have:

    $\displaystyle \frac{1}{3} < x < \frac{7}{3}$
    Last edited by greg1313; Sep 18th 2018 at 12:27 PM.
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  5. #5
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    Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Quote Originally Posted by alexcordero View Post
    Hello all, I think I almost have this but I need to fine tune something...
    Here's the problem: $\displaystyle |-3x+4|-4<-1$
    Here is an advanced easy solution.
    Because it is true that $(\forall a) :|a|=|-a|$ we can rewrite the question.
    $\begin{array}{l} |3x - 4| < 3\\ - 3 < 3x - 4 < 3\\ 1 < 3x < 7\\ \frac{1}{3} < x < \frac{7}{3} \end{array}$
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  6. #6
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    Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Yes your work and answer are correct and you rearranged it correctly. But the solution [1/3, 7/3] is not acceptable, your solution should be (1/3, 7/3) to show that 1/3 and 7/3 are not included in the solution set.
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    Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

    Quote Originally Posted by melis View Post
    Yes your work ------> No, that is false. See * below.

    and answer are correct ------> No, not only is that false, you contradict yourself at the end of your post. See ** below.

    and you rearranged it correctly. But the solution [1/3, 7/3] is not acceptable, your solution should be (1/3, 7/3) to show that 1/3 and 7/3 are not included in the solution set.
    * This step written by the OP is wrong:

    2. $\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}$

    The direction of the inequality symbols reverse at the same moment that sides are multiplied or divided by a negative number.
    It is an incorrect step.

    It has to be this instead:

    $\displaystyle \frac{-7}{-3}>\frac{-3}{-3}x >\frac{-1}{-3}$


    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


    ** You stated "... and answer are correct."

    "[1/3, 7/3] is not acceptable, " as you stated, because it is the wrong answer. It gets marked wrong.
    Last edited by greg1313; Sep 20th 2018 at 01:16 PM.
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