SOLUTION: Find four consecutive even integers such that the sum of twice the fourth integer and half the second integer is 68.

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Question 158470: Find four consecutive even integers such that the sum of twice the fourth integer and half the second integer is 68.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
X, X+2, X+4, X+6 ARE THE 4 EVEN INTEGERS.
2(X+6)+(X+2)/2=68
2X+12+(X+2)/2=68
2X+(X+2)/2=68-12
(2*2X+X+2)/2=56
(4X+X+2)/2=56
5X+2=2*56
5X=112-2
5X=110
X=110/5
X=22 THE FIRST INTEGER.
PROOF:
2(22+6)+(22+2)/2=68
2*28+24/2=68
56+12=68
68=68