Upper bound for $|3x - 2|$ given $|x+3| < 2$?

Question

I am given a condition $|x+3| < 2$ to find the upper bound of $|3x - 2|$. I am not quite sure if my process is correct. \begin{align*} |x+3|<2 & \Longleftrightarrow -2 < x + 3 < 2\\\\ & \Longleftrightarrow -6 < 3x + 9 < 6\\\\ & \Longleftrightarrow -17 < 3x - 2 < -5 \end{align*}

Is it safe to assume that the upper bound of $|3x - 2|$ is $-5$?

Answer 1

No, the upper bound is actually 17 because when you take the absolute value it makes it positive. Also, the RHS of your inequalities should be -5 instead of 5.

Answer 2

Here it is another way to approach it for the sake of curiosity: \begin{align*} |3x - 2| = |3(x + 3) - 9 - 2| = |3(x+3) - 11| \leq 3|x+3| + |11| < 3\times 2 + 11 = 17 \end{align*}

Hopefully this helps!

Answer 3

Continuing your argument:

$-17 < 3x - 2 < -5 \implies 3x-2<0 \implies |3x-2|=-(3x-2)$

$-17 < 3x - 2 < -5 \implies 17 > -(3x - 2) > 5 \implies 5 < |3x - 2| < 17 $