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Math Help - Solve the inequality

  1. #1
    Junior Member
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    Solve the inequality

    Solve the inequality
    x-5/3-x less than 0






    ok this is what i got

    x-5=0
    x=5

    3-x=0
    x=3

    my test intervals were (infinity, 5) (5,3) (3,infinity)

    what is my solution set???
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  2. #2
    Super Member Bacterius's Avatar
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    \frac{x - 5}{3 - x} \leq 0

    - If 3 - x < 0 eg. -x < -3 eg. x > 3, then the inequality is reversed :

    x - 5 \geq 0
    x \geq 5

    - If 3 - x > 0 eg. -x > -3 eg. x < 3, then the inequality is not reversed :

    x - 5 \leq 0
    x \leq 5

    Therefore, your fraction will be lesser than 0 for x \geq 5 and x < 3.
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  3. #3
    Grand Panjandrum
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    How bacterius' post should have looked:

    Your inequality is:

    \frac{x-5}{3-x}\le 0

    Case 1
    If 3-x>0 (which is equlvalent to x<3) the inequality becomes:

     <br />
x-5\le 0<br />

    or

    x \le 5

    So the inequalityholds for x<3


    Case 2
    If 3-x<0 (which is equlvalent to x>3) the inequality becomes:

     <br />
x-5\ge 0<br />

    or x \ge 5

    So the inequality holds for x \ge 5

    Combining these two cases we have the inequality holds when x<3 or x \ge 5

    CB
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  4. #4
    MHF Contributor

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    Quote Originally Posted by flexus View Post
    Solve the inequality
    x-5/3-x less than 0






    ok this is what i got

    x-5=0
    x=5

    3-x=0
    x=3

    my test intervals were (infinity, 5) (5,3) (3,infinity)

    what is my solution set???
    (5,3)? Surely you know that 3< 5! Your test intervals are (-infinity, 3), (3, 5), and (5, infinity). Now just test one number in each interval.

    0 is in (-infinity, 3) and (0- 5)/(3- 0)= -5/3< 0.

    4 is in (3, 5) and (4-5)/(3- 4)= -1/-1= 1> 0

    6 is in (5, infinity) and (6- 5)/(3- 6)= 1/-3= -1/3< 0.
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  5. #5
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    so what is my solution set???
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  6. #6
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    Hello flexus
    Quote Originally Posted by flexus View Post
    so what is my solution set???
    The solution set is
    (-\infty,3)\cup(5,+\infty)
    if the original inequality is
    \frac{x-5}{3-x}<0
    and
    (-\infty,3)\cup[5,+\infty)
    if (as earlier posts have assumed) it's
    \frac{x-5}{3-x}\le0
    Grandad
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