Absolute-Value Inequality: -2|3x + 6| < 4

[SCSmith]

-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.

 

[opusman]

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?

 

[soroban]

Hello, SCSmith!

A variation of opusman's solution . . .

\(\displaystyle \,-2|3x\,+\,6|\; <\, 4\)

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: \(\displaystyle \,-\infty\,<\,x\,<\,\infty\)