Find three consecutive integers such that the sum of one-half of the first plus one-third of the second plus one-fourth of the third is 10.

Answer 1

x-1, x, x+2, what is x??

(x-1)/2 + x/3 + (x+1)/4 = 10

multiply by 12

6x-6 +4x + 3x+3 = 120

13x = 120-9 = 111

x = 111/13 = 8 7/8

x+1 = 9 7/8

x-1 = 7 7/8

there are no 3 consecutive integers who satisfy the required sum

8/2 +9/3 +10/4 = 4+3+2.5 = 9.5 is very close though and = 10 if you round up

9/2 +10/3 +11/4 = 4.5 +3.333.. + 2.75 =10.58333... is also close but sum is too high

maybe go with 8, 9, and 10