Respuesta :

tramserran
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