SOLUTION: find four consecutive even integers where three times the first is ten more than twice the last

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Question 351462: find four consecutive even integers where three times the first is ten more than twice the last
Answer by Fombitz(32388) About Me  (Show Source):
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Let N, N%2B2, N%2B4, and N%2B6 be the consecutive even integers.
3N=10%2B2%2A%28N%2B6%29
3N=10%2B2N%2B12
3N=2N%2B22
N=22
The integers are 22,24,26,and 28.