Results 1 to 6 of 6

Thread: Solve the inequality

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    61

    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???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    $\displaystyle \frac{x - 5}{3 - x} \leq 0$

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

    $\displaystyle x - 5 \geq 0$
    $\displaystyle x \geq 5$

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

    $\displaystyle x - 5 \leq 0$
    $\displaystyle x \leq 5$

    Therefore, your fraction will be lesser than $\displaystyle 0$ for $\displaystyle x \geq 5$ and $\displaystyle x < 3$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    How bacterius' post should have looked:

    Your inequality is:

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

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

    $\displaystyle
    x-5\le 0
    $

    or

    $\displaystyle x \le 5$

    So the inequalityholds for $\displaystyle x<3$


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

    $\displaystyle
    x-5\ge 0
    $

    or $\displaystyle x \ge 5$

    So the inequality holds for $\displaystyle x \ge 5$

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

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,795
    Thanks
    3035
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2009
    Posts
    61
    so what is my solution set???
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello flexus
    Quote Originally Posted by flexus View Post
    so what is my solution set???
    The solution set is
    $\displaystyle (-\infty,3)\cup(5,+\infty)$
    if the original inequality is
    $\displaystyle \frac{x-5}{3-x}<0$
    and
    $\displaystyle (-\infty,3)\cup[5,+\infty)$
    if (as earlier posts have assumed) it's
    $\displaystyle \frac{x-5}{3-x}\le0$
    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solve inequality.
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Sep 1st 2011, 04:06 AM
  2. How to solve this inequality?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Sep 17th 2009, 07:50 AM
  3. Solve the inequality
    Posted in the Algebra Forum
    Replies: 10
    Last Post: May 13th 2008, 05:18 PM
  4. solve inequality
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Dec 22nd 2007, 11:35 AM
  5. When? does the inequality solve
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 1st 2007, 12:54 PM

Search Tags


/mathhelpforum @mathhelpforum