SOLUTION: how to find 3 consecutive even integers such as that 3 times the sum of the last two even integers is 40 more than 5 times the first integer

Question 951318: how to find 3 consecutive even integers such as that 3 times the sum of the last two even integers is 40 more than 5 times the first integer
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
n=first integer; n+2=second integer; n+2=third integer
5n+40=3((n+2)+(n+4))
5n+40=6n+18 Subtract 18 from each side.
5n+22=6n Subtract 5n from each side.
22=n ANSWER 1: The first integer is 22.
n+2=22+2=24 ANSWER 2: The second integer is 24.
N+4=22+4=26 ANSWER 3: The third integer is 26.
CHECK:
3 times sum of last 2 is 5 times first plus 40
5(22)+40=3(24+26)
110+40=3(50)
150=150
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