SOLUTION: find three consecutive even integters such that five times the second equals twice the sum of the first and third

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Question 163445: find three consecutive even integters such that five times the second equals twice the sum of the first and third
Found 2 solutions by mathispowerful, checkley77:
Answer by mathispowerful(115) About Me  (Show Source):
You can put this solution on YOUR website!
find three consecutive even integters such that five times the second equals twice the sum of the first and third
Let the three integers be: x, x+2, x+4
then 5(x+2) = 2[x+(x+4)]
solve it: 5x + 10 = 2(2x + 4)
simplify 5x + 10 = 4x + 8
5x -4x = 8 - 10
x = -2
so the three integers are -2, 0, 2

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
x, x+2, x+4 are the 3 integers.
5(x+2)=2(x+x+4)
5x+10=2(2x+4)
5x+10=4x+8
5x-4x=8-10
x=-2 the first integer.
-2+2=0 for the second integer.
-2+4=2 for the largest integer.
Proof:
5(0)=2(-2+2)
0=2*0
0=0