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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(6x-is 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
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Quote:
 Originally Posted by okayzee1 5x-(4x=3)+2(6x-is greater than or equal to 10x is greater than 4x+4
I see an error.

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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