SOLUTION: Find three consecutive odd integers such that twice the sum of the first and second is one less than three times the third.

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Question 1058182: Find three consecutive odd integers such that twice the sum of the first and second is one less than three times the third.
Found 2 solutions by math_helper, Boreal:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Let:
n = smallest of the 3 odd numbers
n + 2 = 2nd consecutive odd number
n + 4 = 3rd consecutive odd number
"Twice the sum of the 1st and 2nd" is
2((n)+(n+2))
"One less than 3 times the 3rd" is
3(n+4) - 1
—
Set the two equal:
+2%28%28n%29+%2B+%28n%2B2%29%29+=+3%28n%2B4%29-1+
++4n+%2B+4+=+3n+%2B+12+-+1+
+++n+=+7+
—
Ans: The three odd numbers are 7, 9, and 11
—
Check:
2(7+9) = 32
3(11) - 1 = 32 (ok)


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
They are x, x+2, and x+4
2(x+x+2)+1=3(x+4)
2(2x+2)+1=3x+12
4x+5=3x+12
x=7
x+2=9
x+4=11
2(16)+1=33
3*11=33
7,9,11