SOLUTION: 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

Question 53379This question is from textbook Advanced Mathmatics
: 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 This question is from textbook Advanced Mathmatics

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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

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