Let P = {x | 4 ≤ x ≤ 10 and x is even}. Then P is equivalent to:

A set given in set-builder form, {x : conditions on x}, is defined by testing each candidate value of x against every stated condition; only values satisfying all the conditions jointly belong to the set. When two conditions are joined by 'and', a value must pass both to qualify.

1. The two conditions on x are: 4 ≤ x ≤ 10 (x lies in the closed interval from 4 to 10) and x is even.

2. List every integer that satisfies the interval condition alone: 4, 5, 6, 7, 8, 9, 10.

3. From this list, retain only the integers that also satisfy the even condition: 4, 6, 8, 10.

4. Collect the surviving values in roster (list) notation to obtain P.

• Endpoint check: since the inequality uses ≤ (inclusive), both endpoints 4 and 10 qualify under the interval condition, and both are even, so both remain in P.

• Exclusion check: 5 lies inside the interval but is odd, so the even condition removes it; 12 is even but lies outside the interval, so the interval condition removes it — both correctly stay out of P.

Therefore, P = {4, 6, 8, 10}.