Given that a and b are real numbers and that x satisfies the inequality ax>=b when x>=75, what is the smallest integer x that satisfies ax>a+2b ?

To find the smallest integer x that satisfies the inequality ax > a + 2b, we need to consider the given inequality ax ≥ b when x ≥ 75.

Since we don't have specific values for a and b, we need to solve this problem in a more general way.

1. Let's start by considering the two inequalities:
- ax ≥ b (Given inequality)
- ax > a + 2b (Inequality to be satisfied)

2. We want to find the smallest integer x that satisfies both inequalities. This means that x needs to satisfy both ax ≥ b and ax > a + 2b.

3. Let's consider the given inequality ax ≥ b when x ≥ 75. Since we want the smallest x that satisfies both inequalities, we need to find the minimum value of x from ax ≥ b.

4. Divide both sides of the inequality by a (assuming a ≠ 0):
x ≥ b/a

5. Since we know x needs to be at least 75, we can set the value of x as x = max(b/a, 75). This ensures that x satisfies the given inequality ax ≥ b when x ≥ 75.

6. Now, we need to check if x = max(b/a, 75) also satisfies the inequality ax > a + 2b.

7. Substitute x = max(b/a, 75) into the inequality and simplify:
a(max(b/a, 75)) > a + 2b

8. Expand the left-hand side:
max(b, 75a) > a + 2b

9. Now we have two cases:
a) If b > 75a, then the smallest integer x that satisfies both inequalities is x = 75 (since b/a < 75).
b) If b ≤ 75a, then the smallest integer x that satisfies both inequalities is x = ceil(b/a), where ceil(b/a) represents the smallest integer greater than or equal to b/a.

In summary, the smallest integer x that satisfies the inequality ax > a + 2b is:
- If b > 75a, then x = 75.
- If b ≤ 75a, then x = ceil(b/a).