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Math Help - Absolute Values

  1. #1
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    Absolute Values

    I am curious how to proceed with this problem:

    |(5/7)x+(1/4)|<3/4

    I know I must remove the Absolute value, but how?

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Zinefer View Post
    I am curious how to proceed with this problem:

    |(5/7)x+(1/4)|<3/4

    I know I must remove the Absolute value, but how?

    Thanks in advance.
    Hello Zinefer,

    \left \vert \frac{5x}{7}+\frac{1}{4} \right \vert \ < \frac{3}{4}

    Remember this: \vert a \vert < b means -b < a < b. So,

    -\frac{3}{4}<\frac{5x}{7}+\frac{1}{4}<\frac{3}{4}

    How's that for a start?
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  3. #3
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    -\frac{3}{4}<\frac{27x}{28}<\frac{3}{4}

    Would that be the answer?

    Again, thanks for your time.
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  4. #4
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    Not exactly, go bact to this step:

    -\frac{3}{4}<\frac{5x}{7}+\frac{1}{4}<\frac{3}{4}


    You will want to solve for x.

    Multiply each term by 28 to remove the fractions. Then subtract 7 from both sides and the middle.

    Code:
     
    -21 < 20x + 7 < 21
     -7       - 7   -7
    ---------------------
    -28 < 20x     < 14
    Now, divide each term by 20, reduce your fractions, and you will find the range of x.
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  5. #5
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    Thanks alot. The answer is:

    -\frac{7}{5} < x < \frac{7}{10}

    (-\infty, -\frac{7}{5}) \cup (\frac{7}{10},\infty)
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  6. #6
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    Quote Originally Posted by Zinefer View Post
    Thanks alot. The answer is:

    -\frac{7}{5} < x < \frac{7}{10}
    Correct
    Quote Originally Posted by Zinefer View Post
    (-\infty, -\frac{7}{5}) \cup (\frac{7}{10},\infty)
    Actually x is the set of points in the interval \left(-\frac{7}{5}, \frac{7}{10}\right).

    The solution set could also be written \left \{ x \vert -\frac{7}{5}< x < \frac{7}{10}\right\}
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