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Math Help - Rearrange The Inequality

  1. #1
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    Rearrange The Inequality


    Can someone please help me with this problem, I can't seem to figure it out. Thank you in advance.
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  2. #2
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    8/x +1/(x+1) < 9/(x+2)

    8/x +1/(x+1) -9/(x+2) < 0

    Combine the 3 fractions into one fraction only,
    the common denominator is the product of the 3 denominators,

    [8(x+1)(x+2) +1(x)(x+2) -9(x)(x+1)] / [x(x+1)(x+2)] < 0

    [8(x^2 +3x +2) +x^2 +2x -9(x^2 +x)] / [x(x+1)(x+2)] < 0

    [8x^2 +24x +16 +x^2 +2x -9x^2 -9x] / [x(x+1)(x+2)] < 0

    [17x +16] / [x(x+1)(x+2)] < 0

    Therefore,
    A = 17
    B = 16


    The interval where the inequality is defined:
    Solve for x.

    17x +16 < 0
    17x < -16
    x < -16/17 ---no reversing of sense because it dividing by positive 17 only.

    Therefore, the interval is (-infinity,-16/17)
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ticbol View Post
    8/x +1/(x+1) < 9/(x+2)

    8/x +1/(x+1) -9/(x+2) < 0

    Combine the 3 fractions into one fraction only,
    the common denominator is the product of the 3 denominators,

    [8(x+1)(x+2) +1(x)(x+2) -9(x)(x+1)] / [x(x+1)(x+2)] < 0

    [8(x^2 +3x +2) +x^2 +2x -9(x^2 +x)] / [x(x+1)(x+2)] < 0

    [8x^2 +24x +16 +x^2 +2x -9x^2 -9x] / [x(x+1)(x+2)] < 0

    [17x +16] / [x(x+1)(x+2)] < 0

    Therefore,
    A = 17
    B = 16


    The interval where the inequality is defined:
    Solve for x.

    17x +16 < 0
    17x < -16
    x < -16/17 ---no reversing of sense because it dividing by positive 17 only.

    Therefore, the interval is (-infinity,-16/17)
    So far so good, but ticbol missed a few intervals in the solution.

    You want to find "critical points" for the solution. These are points where the numerator or denominator are 0. So we have x = -2, -1, -16/17, and 0.

    Now you want to break the real line into intervals according to these critical points and test each interval in the inequality:
    ( -\infty, -2);~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} > 0 No.

    ( -2, -1);~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} < 0 Yes.

    \left ( -1, -\frac{16}{17} \right );~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} > 0 No.

    \left ( -\frac{16}{17}, 0 \right );~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} < 0 Yes.

    ( 0, \infty );~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} > 0 No.

    So the solution set is the union of the two intervals:
    ( -2, -1) ,  ~ \left ( -\frac{16}{17}, 0 \right )

    -Dan
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  4. #4
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    Quote Originally Posted by DesiKid89 View Post

    Can someone please help me with this problem, I can't seem to figure it out. Thank you in advance.
    I attach a PDF with another solution, 'cause LaTeX's line doesn't support more than 400 characters.
    Attached Files Attached Files
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  5. #5
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    Nice, thanks guys. I got the question right, it looks so hard when you first look at it, but then when you understand how to do it it's like, "Why was I having difficulty with this before?" lol. But once again, I really appreciate the help, thank you.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DesiKid89 View Post
    Nice, thanks guys. I got the question right, it looks so hard when you first look at it, but then when you understand how to do it it's like, "Why was I having difficulty with this before?" lol. But once again, I really appreciate the help, thank you.
    Don't worry about it. Some things just have to be demonstrated before they click. The point is that the next time you see one of these you'll know what to do.

    -Dan
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