S SCSmith New member Joined Oct 25, 2005 Messages 29 Feb 25, 2007 #1 -2|3x + 6| < 4 -4 < 2(3x + 6) < 4 -4 < 6x +12 < 4 Distributive Prop. -16 < 6x < -8 Subtract 12 from each side. -16/6 < x < -8/6 Reduce -8/3 < x <-4/3 This isn't the right solution.
-2|3x + 6| < 4 -4 < 2(3x + 6) < 4 -4 < 6x +12 < 4 Distributive Prop. -16 < 6x < -8 Subtract 12 from each side. -16/6 < x < -8/6 Reduce -8/3 < x <-4/3 This isn't the right solution.
O opusman New member Joined Feb 16, 2007 Messages 4 Feb 25, 2007 #2 think about it, you are taking the absolute value of a number, which is always positve, then multiplying by -2, what type of number will always be?
think about it, you are taking the absolute value of a number, which is always positve, then multiplying by -2, what type of number will always be?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Feb 25, 2007 #3 Hello, SCSmith! A variation of opusman's solution . . . \(\displaystyle \,-2|3x\,+\,6|\; <\, 4\) Click to expand... Divide both sides by \(\displaystyle \,-2\) and reverse the inequality: \(\displaystyle \:|3x\,+\,6|\;> \:-2\) It says: "The absolute value of (something) is greater than negative two." Hey! .The absolute value of ANYTHING is at least zero. . . So, \(\displaystyle 3x\,+\,6\) can be "anything" Therefore, \(\displaystyle x\) can be any real number: \(\displaystyle \,-\infty\,<\,x\,<\,\infty\)
Hello, SCSmith! A variation of opusman's solution . . . \(\displaystyle \,-2|3x\,+\,6|\; <\, 4\) Click to expand... Divide both sides by \(\displaystyle \,-2\) and reverse the inequality: \(\displaystyle \:|3x\,+\,6|\;> \:-2\) It says: "The absolute value of (something) is greater than negative two." Hey! .The absolute value of ANYTHING is at least zero. . . So, \(\displaystyle 3x\,+\,6\) can be "anything" Therefore, \(\displaystyle x\) can be any real number: \(\displaystyle \,-\infty\,<\,x\,<\,\infty\)
