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Math Help - How to solve |2x/3 + 5| ≥ 6 ??

  1. #1
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    Question How to solve |2x/3 + 5| ≥ 6 ??

    |2x/3 + 5| ≥ 6


    I cant figure out how to solve this one...I tried (im not sure if it's correct) andi couldnt finish it off.

    Thankssss in advance :-)
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Cherry View Post
    |2x/3 + 5| ≥ 6


    I cant figure out how to solve this one...I tried (im not sure if it's corrent) but i couldnt finish it off.

    Thankssss in advance :-)
    when you have an absolute value inequality you have to realize that |-6|=6

    so that when you have an inequality like |x-2|>3 you can solve by rewriting it to be: x-2>3 and/or x-2<-3 thus x would be x>5 and/or x<1

    now you try solving your inequality and show us the work.
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    Quote Originally Posted by Quick View Post
    when you have an absolute value inequality you have to realize that |-6|=6

    so that when you have an inequality like |x-2|>3 you can solve by rewriting it to be: x-2>3 and/or x-2<-3 thus x would be x>5 and/or x<1

    now you try solving your inequality and show us the work.
    What Quick said works here, but there are times such that does not work, here is a problem I posed on the forums because of the unsual solution:
    |x|+|x+1|=0
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  4. #4
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    okay thanks guys..

    here i tried to solve it...i hope its correct...

    |2x/3 + 5| ≥ 6

    2x + 15 divided by 3 ≥ 6

    2x + 5 ≥ 6

    2x ≥ 6 - 5

    x ≥ 1/2

    or
    |2x/3 + 5| ≥ -6

    = x ≥ -11/2

    gosh...why do i feel like its wrong?? anyhow please let me know if its correct or wrong thanks!
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    Quote Originally Posted by Cherry View Post
    okay thanks guys..

    here i tried to solve it...i hope its correct...

    |2x/3 + 5| ≥ 6

    2x + 15 divided by 3 ≥ 6
    But you divide 15 by 3 but not 2, you "canceled out of a sum". In fact, you combined fractions from the first step and then uncombined fractions in the second step, why? In the second line you should have multiplied both sides by 3 thus,
    |2x+15| ≥ 18
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  6. #6
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    Oh oh how stupid of me..I cancelled out the sum!!

    so..

    |2x+15| ≥ 18

    2x ≥ 18 - 15

    2x ≥3

    x ≥ 3/2

    it's correct, right?

    Thanks ThePerfectHacker. I didnt pay attention to the fact that 3 is divided by 2x + 15 not only 15.
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    Quote Originally Posted by Cherry View Post
    Oh oh how stupid of me..I cancelled out the sum!!

    so..

    |2x+15| ≥ 18

    2x ≥ 18 - 15

    2x ≥3

    x ≥ 3/2

    it's correct, right?

    Thanks ThePerfectHacker. I didnt pay attention to the fact that 3 is divided by 2x + 15 not only 15.
    Not exactly you forgot the thing Quick said to add a plus minus in from of the 18 (I said it is usually wrong method but in high schoold problems it works) I see no plus-minus sign where you are working.
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    Yeah I know, I figured out how to do the other one and yep you're right I better add the signs.

    Thanks again, and thanks to Quick. You guys are so much better than my teacher who perfers to ignore than help out pfft. You guys rock
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  9. #9
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    What Quick said works here, but there are times such that does not work, here is a problem I posed on the forums because of the unsual solution:
    |x|+|x+1|=0
    How do you solve for x in that problem (in fact I can't see how two positive numbers form 0)?
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  10. #10
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    Quote Originally Posted by Quick View Post
    How do you solve for x in that problem (in fact I can't see how two positive numbers form 0)?
    It is a question that tests your understanding of math. I dare you to ask it to your stupid physics teacher no way she can get it.
    Did you try graphing them?
    y=abs(x) and y=abx(x+1) and see where they interesect.
    ---
    The solution to that equation (and this is why I like it so much besides for it being my problem) is that,
    -1<=x<=0
    There are infinitely many solutions!
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  11. #11
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The solution to that equation (and this is why I like it so much besides for it being my problem) is that,
    -1<=x<=0
    There are infinitely many solutions!
    |x|+|x+1|=0

    substitute: |-1|+|-1+1|=0

    thus: 1=0


    as said before, I don't think there is any possible value for x
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  12. #12
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    I actually meant to say,
    |x|+|x+1|=1
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  13. #13
    TD!
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    As a reference, ThePerfectHacker posted this problem here, where you can also find a solution.
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  14. #14
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    Quote Originally Posted by Cherry View Post
    |2x/3 + 5| ≥ 6


    I cant figure out how to solve this one...I tried (im not sure if it's correct) andi couldnt finish it off.

    Thankssss in advance :-)
    Ok,here I am giving u the solution of this.
    <br />
|2x/3 + 5| ≥ 6<br />
So,(2x/3+5)≥6                                        or,-(2x/3+5)≥6<br />
or,2x/3≥6-5                                               or,(2x/3+5)<=-6<br />
or,2x/3 ≥1                                                  or,2x/3<=-11<br />
or,x≥3/2                                                     or,x<=-33/2<br />
           So,the result is x>=3/2 or,x<=-33/2
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  15. #15
    Junior Member qspeechc's Avatar
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    |2x/3+5| ≥ 6

    I think it is important to remember:

    |x| = x if x ≥ o
    and
    |x| = -x if x 0

    So:
    If 2x/3 + 5 ≥ 0, then:

    2x/3+ 5 ≥ 6

    And solve:
    x ≥ 3/2

    Now, if 2x/3 + 5 ≤ 0, then:

    -(2x/3 + 5) ≥ 6

    2x/3 + 5 -6

    2x/3 -11

    x ≤ -33/2
    Last edited by qspeechc; November 10th 2007 at 07:30 AM.
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