How many possible integer values are there for x if |4x-3| < 6 ?

A. ONE
B. TWO
C. THREE
D. FOUR
E. FIVE

|4x - 3| < 6
=> -6 < 4x -3 < 6
=> -3 < 4x < 9
=> -3/4 < x < 9/4
Hence integer values x can take are : 0, 1, 2
Hence Ans C

|4x-3| < 6 implies -6 < 4x-3 < 6
Taking these individually:
4x - 3 < 6 --------> 4x < 9 --------> x < 9/4 --------> x <= 2 (since x is integer).
4x - 3) > -6 --------> 4x > -3 --------> x > -3/4 --------> x >= 0 (since x is integer).

So combining these two we can say x = 0, 1 or 2. So answer should be C.

I really like akdayal's solution above. Below is an alternative:

In word translations, the absolute value of a difference of two terms means the distance between the two terms on the number line: |4x-3| means the distance between 4x and 3 on the number line. Thus we can rephrase the prompt to say "The distance between 3 and 4x is less than 6". On a number line, this means that 4x can be any value that is less than 6 units away from 3. So 4x can be anything between -3 and 9 (these are the boundaries of values 6 units away from 3).
Furthermore, since we're only interested in integers, we only need to consider 4x = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, or 8}. Of these values, the only ones that would result in integer x are:

• 4x=0 -> x=0

• 4x=4 -> x=1

• 4x=8 -> x=2

. Thus there can only be 3 possible integer values for x. The answer is C.

To practice similar questions, use the Drill Engine to generate timed drills and set topic='Inequalities & Absolute Values' and difficulty='700+'

-Patrick