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Math Help - Inequalities help!

  1. #1
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    Inequalities help!

    Solve the ff inequalities:
    need solution set
    4.) 2 > -3 - 3x >= -7
    5.) 4/x - 3 - 2/x + 7 > 0
    6.) 1/(x+1) < 2/(3x - 1)
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    2 > -3 - 3x >= -7
    If what I'm about to do bothers you, please feel free to do each step on the two inequalities:
    2 > -3 - 3x and -3 - 3x >= 7

    So
    2 > -3 - 3x >= -7

    Add 3 to all three "sides":
    2 + 3 > 3 - 3 - 3x >= 3 - 7

    5 > -3x >= -4

    Divide all three "sides" by -3. Since we are dividing by a negative quantity, don't forget to switch the > to < and the >= to <=!
    5/(-3) < -3x/(-3) <= -4/(-3)

    -5/3 < x <= 4/3

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    4/x - 3 - 2/x + 7 > 0
    4/x - 3 - 2/x + 7 > 0

    Simplify the LHS:
    2/x + 4 > 0

    Add the two numbers on the LHS:
    2/x + 4*x/x > 0

    (2 + 4x)/x > 0

    Now we wish to look for "critical points." The denominator is equal to zero at x = 0, so this is a cp. Next, the numerator is zero at x = -1/2, so this is a cp.

    Break the real number line into the following pieces:
    (-infinity, -1/2), (-1/2, 0), and (0, infinity)

    For which of these three regions is the inequality satisfied?
    (-infinity, -1/2): (2 + 4x)/x > 0 (Check!)
    (-1/2, 0) : (2 + 4x)/x < 0 (No.)
    (0, infinity): (2 + 4x)/x > 0 (Check!)
    (If we have <= and/or >= we also need to check the cp's themselves.)

    So the solution set is x < -1/2 and x > 0.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    1/(x+1) < 2/(3x - 1)
    1/(x+1) < 2/(3x-1)

    This uses the same method as the previous problem.

    1/(x+1) < 2/(3x-1)

    1/(x+1) - 2/(3x-1) < 0

    Add the fractions:
    [(3x-1) - 2(x+1)]/[(x+1)(3x-1)] < 0

    (x-3)/[(x+1)(3x-1)] < 0

    This has cp's at x = 3, x = 1/3, and x = -1.
    (-infinity, -1): (x-3)/[(x+1)(3x-1)] < 0 (Check!)
    (-1, 1/3): (x-3)/[(x+1)(3x-1)] > 0 (No.)
    (1/3, 3): (x-3)/[(x+1)(3x-1)] < 0 (Check!)
    (3, infinity): (x-3)/[(x+1)(3x-1)] > 0 (No.)

    Thus the solution set is: x < -1 and 1/3 < x < 3.

    Now, if we did this problem the intuitive way:
    1/(x+1) < 2/(3x-1)

    Multiply both sides by (x+1):
    1 < 2(x+1)/(3x-1)

    Multiply both sides by (3x-1):
    3x - 1 < 2(x + 1)
    3x - 1 < 2x + 2
    x < 3

    Note that this did not give the correct answer. This is because we are not saying whether or not (x+1) or (3x-1) is negative when we multiply by them. Always be careful of this!

    -Dan
    Last edited by topsquark; October 6th 2006 at 08:47 AM.
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