Question 192254
Find three consecutive odd integers such that three time the middle integer is seven more than the sum of the first and third integers. 
my try: 
1st integer: n 
2nd integer: 3(n+2) ---- it's n+2
3rd integer: (n+4) + 7 ----- it's n+4
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3*(n+2) = n + n+4 + 7 ---- 3 times the middle # is the sum of 2 #s + 7
3n+6 = 2n+11
n = 5
5, 7, 9
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n+3(n+2) = 7+(n+4) 
2n + 5 = 11 + n 
n + 5 = 11 
n = 6