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Solving two absolute value variables.

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Given: |A|+A+B=15 and A+|B|-B=13. What is A+B equal to? Give all possibilities.

    2. Relevant equations



    3. The attempt at a solution

    I solve for both absolute variables. So,

    A=15-A-B or A=A+B-15
    and
    B=13-A+B or B=A-B-13

    Firstly, I solve for A.

    A+A=15-B
    2A=15-B
    A=[tex]\frac{15-B}{2}[/tex]

    and
    A-A=B-15
    0=B-15
    B=15

    Plug them back in to the original eq. to see if it works.
    |[tex]\frac{15-B}{2}[/tex]|+[tex]\frac{15-B}{2}[/tex]+B=15
    |[tex]\frac{15-B}{2}[/tex]|=15-B-[tex]\frac{15-B}{2}[/tex]
    |[tex]\frac{15-B}{2}[/tex]|=+/- [tex]\frac{15-B}{2}[/tex]
    So that works. But when I plug in B=15 into the original eq. it doesn't.

    So far, I have A=[tex]\frac{15-B}{2}[/tex]. Next, I solve for B.

    B-B=13-A
    0=13-A
    A=13

    and
    B+B=A-13
    2B=A-13
    B=[tex]\frac{A-13}{2}[/tex]

    So, I plug in B into the original equation(the second one given).
    Plugging in A=13 just comes out to B=+/- B.
    Plugging in B=[tex]\frac{A-13}{2}[/tex] results in
    [tex]\frac{A-13}{2}[/tex]=+/- [tex]\frac{A-13}{2}[/tex].

    So, when the problem asks for A+B, how do I add them? Do I do:

    A+B=[tex]\frac{15-B}{2}[/tex]+[tex]\frac{A-13}{2}[/tex] ?

    Please help! Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 15, 2009 #2

    Mark44

    Staff: Mentor

    I would approach this problem by breaking it up into four cases, to get rid of the absolute values.

    I: Assume A > 0 and B > 0.
    In this case, the equations are 2A + B = 15 and A + B - B = 13. The 2nd equation is equivalent to A = 13. Solving for B, I get B = -11, which is a contradiction with the assumption that B > 0.

    II: Assume A > 0 and B < 0.
    With this assumption |A| = A and |B| = -B.
    This case gives me values for A and B that don't contradict the assumption in this case.

    III: Assume A < 0 and B > 0.
    IV: Assume A < 0 and B < 0.
     
  4. Nov 15, 2009 #3
    So, it seems that II is the only one without a contradiction. So, does that mean I only use A=(15-B)/2 and B=(A-13)/2 ?? And just add them A+B=(15-B)+(A-13) / 2. So, my final answer will be (A-B+2)/2 = A+B?
     
  5. Nov 15, 2009 #4
    Right? (A-B+2)/2 gives me the value of A+B? Nothing else would be the value right? I appreciate your help.
     
  6. Nov 15, 2009 #5

    Mark44

    Staff: Mentor

    No, your final answer should be a number. What values did you get for A and B in the 2nd case?
     
  7. Nov 15, 2009 #6
    Oh, well in that case I got A=43/5 and B=-11/5. Thus, A+B=32/5 ? and that is the only answer right?
     
  8. Nov 15, 2009 #7
    Could someone please help me finish this problem? Mark, thanks for everything!
     
  9. Nov 15, 2009 #8

    Mark44

    Staff: Mentor

    Those are the numbers I get for the 2nd case. The only cases I checked are the first two, so if you're confident in your results for the 3rd and 4th cases, then you've got your answer.
     
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