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tramserranLet 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