How many integer values of x satisfy the equation |x-3|+|x-5|<|x-11|?

Understand the problem
We need to find the number of integer solutions for the inequality |x-3| + |x-5| < |x-11|.

Consider the critical points 3, 5, and 11
Break the inequality into cases based on these points.

Case 1: x < 3
|x-3| = 3-x, |x-5| = 5-x, |x-11| = 11-x
3-x + 5-x < 11-x
8-2x < 11-x
8-2x < 11-x
-2x + x < 11-8
-x < 3
x > -3

Case 2: 3 ≤ x < 5
|x-3| = x-3, |x-5| = 5-x, |x-11| = 11-x
x-3 + 5-x < 11-x
2 < 11-x
x < 9

Case 3: 5 ≤ x < 11
|x-3| = x-3, |x-5| = x-5, |x-11| = 11-x
x-3 + x-5 < 11-x
2x-8 < 11-x
3x < 19
x < 19/3
x < 6.33

Case 4: x ≥ 11
|x-3| = x-3, |x-5| = x-5, |x-11| = x-11
x-3 + x-5 < x-11
2x-8 < x-11
x < -3 (No solutions)

Combine the cases
From cases 1 and 2:
-3 < x < 9

From case 3:
5 ≤ x < 6.33

Find integer values
Combining the valid ranges:
5 ≤ x < 9

Integer values: 5, 6, 7, 8

Conclusion
There are 4 integer values of x that satisfy the inequality.