If p = (y : 2y $\geq$ 6) and Q = (y : y -3 $\leq$ 4), where y is an integer, find p$\cap$Q

Step 1: Analyze the question We need the intersection of two sets of integers, P and Q. The intersection contains only the values that satisfy both conditions.

Step 2: Solve each inequality

• P: 2y ≥ 6, so y ≥ 3
• Q: y − 3 ≤ 4, so y ≤ 7

Step 3: Find the intersection P ∩ Q contains the integers with 3 ≤ y ≤ 7:

P ∩ Q = {3, 4, 5, 6, 7}

Step 4: Check against the options

• A {3, 4}: too few elements
• B {3, 7}: leaves out 4, 5 and 6
• C {3, 4, 5, 6, 7}: matches our result
• D {4, 5, 6}: wrongly excludes the end values 3 and 7

Both inequalities include "equal to" (≥ and ≤), so the end values 3 and 7 are part of the set.

Exam tip: When you see ≥ or ≤, include the boundary values. Whenever the signs are > or <, leave them out. For intersection questions, first write each set as a range, then pick the overlap, which here is 3 ≤ y ≤ 7.