SOLUTION: Find 3 consecutive odd integers such that the sum of the 1st, 2nd and 3 times the 3rd is 58 more than the smaller number?

Question 725993:
Find 3 consecutive odd integers such that the sum of the 1st, 2nd and 3 times the 3rd is 58 more than the smaller number?
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
LET X, X+2 & X+4 BE THE 3 INTEGERS.
X+(X+2)+3(X+4)=X+58
X+X+2+3X+12=X=58
5X-X=58-2-12
4X=44
X=44/4
X=11 ANS. FOR THE SMALLER INTEGER.
11+2=13 ANS. FOR THE MIDDLE NUMBER.
11+4=15 ANS. FOR THE LARGEST INTEGER.
PROOF:
11+13+3*15=11+58
24+45=69
69=69
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