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August 16th, 2009, 07:03 PM  #1 
Newbie Joined: Aug 2009 Posts: 1 Thanks: 0  An impossible compound equality(11th grade)
Please i need help. its the 1st week of school and im lost. I have to solve compound enequality problems that lack and or or. 5x(4x=3)+2(6xis greater than or equal to 10x is greater than 4x+4 then 9x+16 is less than or equal to to 4x+3 is less than or grerater to 22x(2x+6) 
August 16th, 2009, 08:32 PM  #2  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Quote:
Second part will be stated as 9x + 16 ? 4x + 3, where 4x + 3 < 22x  (2x + 6) = 20x  6 or 4x + 3 > 22x  (2x + 6) = 20x  6. For the last two inequalities I have just mentioned, the result will be 9 < 16x or 9 > 16x, therefore x > 9/16 or x < 9/16. For 9x + 16 ? 4x + 3, 5x ? 13, therefore x ? 13/5. This inequality must be hold, so since previously we have determined that x > 9/16 or x < 9/16 can possibly be hold, we can ignore x > 9/16 and worry about x < 9/16. Therefore, logically solving this inequality and x ? 13/5 and the answer for the second part of your post is obviously x ? 13/5.  
August 17th, 2009, 12:04 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,935 Thanks: 2209 
I assume the first problem should be 5x  (4x + 3) + 2(6x  8) ? 10x > 4x + 4, i.e., 13x  19 ? 10x > 4x + 4. You can rewrite that as two inequalities: 13x  19 ? 10x and 10x > 4x + 4, which simplify to 3x ? 19 and 6x > 4, i.e., 3x > 2. As the first inequality implies the second, the solution is x ? 19/3. 

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compound, equality11th, grade, impossible 
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