`3|x| + 2 lt 17`

To solve, isolate the absolute value. To do so, subtract both sides by 2.

`3|x|+2-2lt17-2`

`3|x|lt15`

And divide both sides by 3.

`(3|x|)/3lt15/3`

`|x|lt5`

Next, drop the absolute value sign. Note that when dropping the absolute value sign in equality equation for less than ( `|x|lta ` ), the resulting equation is `-altxlta` .

`-5ltxlt5`

**Hence, the solution to the inequality equation `3|x|+2lt17` is `-5ltxlt5` . In interval notation, the solution is `(-5, 5)` .**