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Math Help - Simple absolute value Q

  1. #1
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    Simple absolute value Q

    The question:
    Solve for x,
    |2x-1|+|x+3| = 5

    What's the best way to attempt questions involving the addition of absolute values? I can consider the case when x is negative, and x is positive, but that seems like an elementary approach. Is there a better way?

    Note: I'm not interested in the solution, I'm interested in the method. Thanks.
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  2. #2
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    I suppose you could always graph the function, see where it crosses the line \displaystyle y = 5...
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  3. #3
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    Why would considering the two cases where x is negative and positive be an elementary approach? Also, even if it is, what's wrong with an elementary approach? By doing this, we get x = 1 and -1. I don't see anything wrong with that. As Prove It has already said, you could graph it, but that seems like more work than is necessary.
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    Quote Originally Posted by rtblue View Post
    Why would considering the two cases where x is negative and positive be an elementary approach? Also, even if it is, what's wrong with an elementary approach? By doing this, we get x = 1 and -1. I don't see anything wrong with that. As Prove It has already said, you could graph it, but that seems like more work than is necessary.
    I just figured there'd be a better way. For instance, with this example:

    |2x + 1| = x + 1

    You could square both sides and apply the quadratic formula to get a solution, instead of looking at negative/positive. Just wondering if there was a similar method.
    Last edited by Glitch; March 26th 2011 at 09:01 PM.
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  5. #5
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    In this case you would probably have to square both sides twice...

    \displaystyle |2x - 1| + |x + 3| = 5

    \displaystyle (|2x-1| + |x+3|)^2 = 5^2

    \displaystyle (2x - 1)^2 + 2|2x-1||x+3| + (x + 3)^2 = 25

    \displaystyle 2|2x-1||x+3| = 25 - (2x-1)^2 - (x+3)^2.

    Squaring both sides again will eliminate the absolute values, but you'll end up with a quartic you have to solve...
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  6. #6
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    Indeed.

    With the graphing method, is there a trick to doing it, or is it a matter of subbing in values to determine the shape?
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  7. #7
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    Draw the graphs of \displaystyle |2x - 1| and \displaystyle |x + 3|, then use addition of ordinates to graph \displaystyle |2x - 1| + |x + 3|.
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    I expected that'd be the case. Thanks again.
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  9. #9
    Junior Member mathfun's Avatar
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    Algebrically, you have to take three cases so that absolute values are eliminated.
    1)x<=-3
    2)-3<x<0.5
    3) x>=0.5

    and consider the accepted solutions
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  10. #10
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    Quote Originally Posted by Glitch View Post
    The question:
    Solve for x,
    |2x-1|+|x+3| = 5

    What's the best way to attempt questions involving the addition of absolute values? I can consider the case when x is negative, and x is positive, but that seems like an elementary approach. Is there a better way?

    Note: I'm not interested in the solution, I'm interested in the method. Thanks.
    Similarly, you could say

    Both 2x-1 and x+3 are \ge\ 0 gives

    |2x-1|+|x+3|=2x-1+x+3=5

    and for that case you must have 2x\ge\ 1 and x\ge\ -3 so 2x\ge\ 1


    Next, 2x-1\ge\ 0 and x+3\le\ 0

    This is not possible as we cannot have 2x-1\ge\ 0 and x\le\ -3


    However, we can have 2x-1\le\ 0 and x\ge\ -3

    In this case you have (1-2x)+x+3=5

    (You must verify that the solution given falls correctly within the required range)


    Finally, if 2x-1 and x+3 are both \le\ 0

    you find that (1-2x)-(x+3)=5 requires x to be above -3
    which causes a contradiction.

    So two of the 4 options are valid.
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