March 17th, 2011, 08:28 AM  #1 
Newbie Joined: Mar 2011 Posts: 2 Thanks: 0  Inequality
I'm having trouble with this inequality: 1/(x5) < 4/5x I'm pretty sure the first step is to do this: 1/(x5)  4/5x < 0 But I'm stuck here. Could anyone nudge me in the right direction? 
March 17th, 2011, 08:44 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond  Re: Inequality 
March 17th, 2011, 09:29 AM  #3 
Newbie Joined: Mar 2011 Posts: 2 Thanks: 0  Re: Inequality
Thank you!

March 18th, 2011, 01:39 AM  #4  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
x + 20 > 0, 5x(x  5) < 0 or x + 20 < 0, 5x(x  5) > 0 For 5x(x  5) < 0: 5x > 0, x  5 < 0 or 5x < 0, x  5 > 0 x > 0, x < 5 or x < 0, x > 5 For 5x(x  5) > 0: 5x > 0, x  5 > 0 or 5x < 0, x  5 < 0 x > 0, x > 5 or x < 0, x < 5 For x + 20 > 0 and x + 20 < 0: x > 20 and x < 20, respectively x > 20, x > 0, x < 5 or x < 0, x > 5 OR x < 20, x > 0, x > 5 or x < 0, x < 5 x < 0, x > 5 is equivalent to 5 < x < 0 x > 0, x > 5 is equivalent to x > 0 x < 0, x < 5 is equivalent to x < 5 x > 20, x > 0, x < 5 or 5 < x < 0 OR x < 20, x > 0 or x < 5 x > 20 and 5 < x < 0 is equivalent to 5 < x < 0 x = 1 does not satisfy x > 0, x < 5 5 < x < 0 OR x < 20, x > 0 or x < 5 x = 1 does not satisfy x < 20 and x > 0 x < 20 and x < 5 is equivalent to x < 20 ? 5 < x < 0 or x < 20  
March 18th, 2011, 03:07 AM  #5  
Newbie Joined: Mar 2011 Posts: 7 Thanks: 0  Re: Quote:
the whole equation by the square of the denominator giving u: then just draw it and determine negative domains  
March 18th, 2011, 03:22 AM  #6 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
Your solution is much easier. Thanks, daniel_an!

March 18th, 2011, 04:30 AM  #7 
Newbie Joined: Mar 2011 Posts: 4 Thanks: 0  Re: Inequality
Wow! Nice solution.


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