SOLUTION: Consider three consecutive odd integers. The sum of the first integer and 2 times the second is 54 more than the third. Find the integers.

Question 1035432: Consider three consecutive odd integers. The sum of the first integer and 2 times the second is 54 more than the third. Find the integers.
Found 3 solutions by jorel555, josgarithmetic, MathTherapy:
Answer by jorel555(1290)   (Show Source): You can put this solution on YOUR website!
Let's call the first integer n. The next two odd integers are n+2, and n+4. Then:
n+2(n+2)-54=(n+4)
n+2n+4-54=n+4
2n=54
n=27
n+2=29
n+4=31!!!!!!
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
n is any integer.
Your expressions for your integers are 2n+1, 2n+3, 2n+5.

The description is the equation .
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Consider three consecutive odd integers. The sum of the first integer and 2 times the second is 54 more than the third. Find the integers.

Let smallest/first integer be F
Then others are: F + 2, and F + 4
We then get: F + 2(F + 2) = F + 4 + 54
F + 2F + 4 = F + 58
3F - F = 58 - 4
2F = 54
F, or smallest/first integer = , or
Other 2:
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