ziyuenlau wrote:

For how many integer values of x, is \(|3x-3|+|2x+8|<15\)?

A. 2

B. 3

C. 4

D. 5

E. 6

\(|3x-3|+|2x+8|<15\)

\(3*|x-1|+2*|x+4|<15\)

We want the values of x such that the sum of "thrice their distance from 1" and "twice their distance from -4" is less than 15.

Let's try to find the point where this distance is equal to 15.

........................ (-4) ...................................... (0) ........... (1) ..........................

The distance between -4 and 1 is 5. Thrice this distance is 15. So at the point x = -4, the sum will be 15. As we move to the right of -4, the sum will reduce (since the twice component will keep increasing). At x = 1, the sum becomes 0 + 2*5 = 10.

What happens when you go to the right of 1? Now the sum starts increasing since the thrice components increasing now.

At x = 2, the sum becomes 3*1 + 2*6 = 15.

To the right of 2, the sum will keep increasing.

So the sum will be less than 15 between -4 and 2. This gives us 5 integer values (-3, -2, -1, 0, 1).

Answer (D)

Mam, my question in when x < = -4, then for if x = -5 , |2x+8| is negative. but we are also using x = -4, then |2x+8| becomes "0". then why we are not considering |2x+8| = 0 for x = -4. I don't know whether I made myself clear to you or not