Let x represent the first even integer: x = 2(k) → y = 2k
Let y represent the second even integer: y = 2(k + 1) → y = 2k + 2
x · y = 48
(2k) · (2k + 2) = 48
4k² + 4k - 48 = 0
4(k² + k - 12) = 0
4 (k + 3)(k - 2) = 0
k = -3, k = 2 (Note: POSITIVE integers so k = -3 is ruled out)
x = 2k → x = 2(2) → x = 4
y = 2(k + 1) → y = 2(2 + 1) → y = 2(3) → y = 6
The smaller number (x) is 4
