Here's the question you clicked on:
AGummiBear55
CHECK MY WORK, PLEASE! Solve: 16 < 3x + 1 < 4 16 < 3x + 1 < 4 16 < 3x + 1 -1 to both sides 15<3x /3 to both sides 5<x 3x + 1 < 4 -1 to both sides 3x < 3 /3 x<1 Answer: 5<x and x<1 Did I do that right? I have a feeling I didn't.
It's wrong. First of all, 16 cannot be less than 4.
Is there a mistake in the question?
??? I thought with these types of problems you have to split them up from 16 < 3x + 1 < 4 to two equations; 16 < 3x +1 and 3x +1<4.