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Match the inequalities with the corresponding statements.

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data

    PROBLEM: Match the inequalities with the corresponding statements.
    INEQUALITIES: 1) |a-5|< 1/3
    2) |a- 1/3|< 5
    STATEMENTS: a) The distance from a to 5 is less than 1/3
    b) a is less than 5 units from 1/3



    3. The attempt at a solution

    I solved for a for both problems and for both problems I am getting 14/3 < a< 16/3.
    My problem is that since I am getting the same solution for both problems, then aren't both the statements (a and b) correct for both of the inequalities? Or am I getting the wrong answers when I solve the inequalities? Please advise if you see the problem. Thanks in advance.
     
  2. jcsd
  3. Jun 30, 2011 #2

    vela

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    It seems you've solved the inequalities incorrectly, but solving them really isn't necessary. You want to be able to interpret the inequalities as written. For example, what does |a-5| correspond to geometrically?
     
  4. Jun 30, 2011 #3

    SammyS

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    Of course, vela is correct, but if you want to check what's wrong with your algebraic method, show how you get the answer for each inequality.
     
  5. Jun 30, 2011 #4
    VELA: Geometrically, |a-5| corresponds to moving a over right 5, correct? But how does that help me interpret the inequality as written? Or am I interpreting it incorrecly. Please advise.

    SAMMY: This is how I worked the first one out algebraically.
    |a-5|< 1/3 has two solutions right? One positive and one negative?
    Solution 1:
    a-5 < 1/3
    a< 1/3 + 5
    a < 16/3

    and a-5 > -1/3
    a > -1/3 + 5
    a > 14/3

    So that gives me 14/3 < a< 16/3.
    And I worked out the second problem with the same steps to get the same answer. I'm confused.
     
  6. Jun 30, 2011 #5

    ehild

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    You made a mistake:
    |a-1/3|<5 means -5<a-1/3<5. Adding 5 to all sides, you get a negative number on the left. You just missed that minus sign.

    ehild
     
  7. Jun 30, 2011 #6

    Mark44

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    No, |a - 5| represents the distance between a and 5. Looking at things in terms of transformations, which you seem to be doing, the graph of y = |x - 5| can be seen as the translation of the graph of y = |x| by 5 units to the right.
     
  8. Jun 30, 2011 #7

    vela

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    No. |a-5| is equal to the distance between a and 5. For example, when a=4 which is a distance of 1 away from 5 on the number line, you get |a-5|=1. Similarly, a=6.5, which is 1.5 more than 5, you get |a-5|=1.5.

    So if you had an inequality like |x-1| < 2, you can interpret that as

    Code (Text):
    |x-1|  =  "the distance between [i]x[/i] and 1"
      <    =  "is less than"
      2    =  "two"
     
  9. Jun 30, 2011 #8
    ahh. just what i needed. thanks to all(:
     
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