Inequalities and intervals

Subtract a from both sides of x + a > b: x > b − a. Adding or subtracting the same number preserves the inequality direction.

For integer bounds a and b, the integers strictly between them are a + 1, a + 2, ..., b − 1. Their count is b − a − 1. Exclude both bounds because both inequalities are strict.

Divide both sides of a × x ≤ b by positive a. The direction stays the same: x ≤ b ÷ a. Equality is included in the solution.

Divide the budget by the ticket price and take the whole-number quotient. The ticket count must be an integer and its total price cannot exceed the budget. Check that one more ticket would be unaffordable.

Subtract a from both sides of x + a > b: x > b − a. Adding or subtracting the same number preserves the inequality direction.

Divide both sides of −a × x < b by negative −a. Dividing by a negative reverses the inequality: x > b ÷ (−a).

For integer bounds a and b, the integers strictly between them are a + 1, a + 2, ..., b − 1. Their count is b − a − 1. Exclude both bounds because both inequalities are strict.

The interval (a, b] means a < x ≤ b. The round bracket excludes a, while the square bracket includes b. Check both conditions: the number must exceed a and be at most b.

“At least” means “greater than or equal to”. If the minimum score is a, write x ≥ a; exactly a points also qualifies.

The interval (a, b] means a < x ≤ b. The round bracket excludes a, while the square bracket includes b. Check both conditions: the number must exceed a and be at most b.