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Math Help - Absolute Values Question?

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    Absolute Values Question?

    ok so i have never been given a quesiton like this and im trying to figure it out for the first time. here it is

    Solve |2x-3/4x+1| = > 2
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jamman790 View Post
    ok so i have never been given a quesiton like this and im trying to figure it out for the first time. here it is

    Solve |2x-3/4x+1| = > 2
    one way to go, is to square both sides. you get

    \frac {(2x - 3)^2}{(4x + 1)^2} \ge 2

    now solve for x. be sure to check if there are any extraneous solutions.


    another approach is to account for the inside of the absolute values to be positive and negative. that is, solve the following two inequalities separately:

    \frac {2x - 3}{4x + 1} \ge 2 and - \frac {2x - 3}{4x + 1} \ge 2
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    It says...show all solutions?...but what does that mean?
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    Quote Originally Posted by Jhevon View Post
    one way to go, is to square both sides. you get

    \frac {(2x - 3)^2}{(4x + 1)^2} \ge 2

    now solve for x. be sure to check if there are any extraneous solutions.


    another approach is to account for the inside of the absolute values to be positive and negative. that is, solve the following two inequalities separately:

    \frac {2x - 3}{4x + 1} \ge 2 and - \frac {2x - 3}{4x + 1} \ge 2
    I believe you'll find if you square both sides then the inequation becomes

    \frac{(2x - 3)^2}{(4x + 1)^2}\geq 4.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prove It View Post
    I believe you'll find if you square both sides then the inequation becomes

    \frac{(2x - 3)^2}{(4x + 1)^2}\geq 4.
    but of course. silly mistake on my part
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