What are three consecutive even integers such that the sum of one-half of the first plus one-third of the second plus one-fourth of the third is equal to 19?

Answer 1

x,x+2, and x+4

x/2 +(x+2)/3 + (x+4)/4 = 19

multiply by 12

6x +4x+8 + 3x+12 = 228

13x = 228 -20 = 208

x = 208/13=16

16, 18 and 20

16/2 + 18/3 + 20/4 = 8+6+5 = 19