Question 53379
find four consecutive even intergers such that six times the sum of the first two exceeds twice the sum of the first and the forurth by 416
:
The 4 consecutive even integers: x, (x+2), (x+4), (x+6)
:
Write it just like it says:
"six times the sum of the first two exceeds twice the sum of the first and the fourth by 416"
:
6*(x + (x+2)) - 2(x + (x+6)) = 416
:
6(2x+2) - 2(2x+6) = 416
:
12x + 12 - (4x + 12) = 416
:
12x + 12 - 4x - 12 = 416
:
            8x = 416
:
             x = 416/8
:
             x = 52 is the 1st digit
:
Check: 
6(52+54) - 2(52+58) =
6(106) - 2(110) =
  636 - 220 = 416