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Math Help - Absolute value Inequality

  1. #1
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    Absolute value Inequality

    Hey all,
    I have the following absolute value inequality;
    |3x| - 1 \le |2x-1|
    which i simplify to:
    |3x| - |2x-1| \le 1
    The critical values are, x = 0 and x = 1/2;
    Not sure how to solve it from now on..
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  2. #2
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    Quote Originally Posted by Oiler View Post
    Hey all,
    I have the following absolute value inequality;
    |3x| - 1 \le |2x-1|
    which i simplify to:
    |3x| - |2x-1| \le 1
    The critical values are, x = 0 and x = 1/2;
    Not sure how to solve it from now on..
    Take a visual approach: Draw the graphs of y = |3x| - 1 and y = |2x - 1| and use them as a guide ....
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  3. #3
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  4. #4
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    I have gone through that example and still cant seem to grasp the concept. Would I in this example consider the case when,
    x \le \frac{1}{2} (which would be false in this case)
    and when,
    \frac{1}{2} > x

    Hmm, no idea..
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  5. #5
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    Quote Originally Posted by Oiler View Post
    I have gone through that example and still cant seem to grasp the concept. Would I in this example consider the case when,
    x \le \frac{1}{2} (which would be false in this case)
    and when,
    \frac{1}{2} > x

    Hmm, no idea..
    Did you try doing what was suggested in post #2?
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  6. #6
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    The question suggests that we use the number line instead of a graph, but the grap has touches the x axes at 2/5 and -2. I cant see how I can obtain those two values from original question..
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  7. #7
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    Quote Originally Posted by Oiler View Post
    The question suggests that we use the number line instead of a graph, but the grap has touches the x axes at 2/5 and -2. I cant see how I can obtain those two values from original question..
    You're meant to look at where the graphs intersect (the x-coordinates of these points are required) and then consider what the equation of the lines are at these intersection points.
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  8. #8
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    Quote Originally Posted by Oiler View Post
    Hey all,
    I have the following absolute value inequality;
    |3x| - 1 \le |2x-1|
    which i simplify to:
    |3x| - |2x-1| \le 1
    We study the cases when x\le0,\,0\le x<\dfrac12 and x>\dfrac12.

    For x\le0 the inequality becomes -3x-(1-2x)\le1 then x\ge -2 so -2\le x\le0 is a solution set.

    For 0\le x<\dfrac12 the inequality becomes 3x-(1-2x)\le1 so x\le\dfrac25 then 0\le x\le\dfrac25 is another solution set.

    For x>\dfrac12 the inequality becomes 3x-(2x-1)\le1 so x\le0 but since the assumption is x>\dfrac12 there's no solution here.

    So the solution set for the original problem is \left[ -2,0 \right]\cup \left[ 0,\dfrac{2}{5} \right]=\left[ -2,\dfrac{2}{5} \right].
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