SOLUTION: the sum of three consecutive odd integers is five less than twice the the smallest of these integers

Question 436109: the sum of three consecutive odd integers is five less than twice the the smallest of these integers
Answer by marilynh(7) (Show Source): You can put this solution on YOUR website!
If the three numbers are consecutive and odd, and we let the first number be represented by x, then the second number would be 2 more than the x and the third number would be 4 more than x. The integers we are looking for are:
x, (x+2), (x+4)
We want the sum of these to equal to "five less than twice the smallest integer" (we represented the smallest integer by x). This gives us the equation
x + (x+2) + (x+4) = 2x - 5
Solve the equation for x:
x + (x+2) + (x+4) = 2x - 5
3x + 6 = 2x - 5
3x - 2x = -5 - 6
1x = -11
The integers we are looking for are -11, -9, -7, found by substituting -11 for x:
x = -11
x + 2 = -11 + 2 = -9
x + 4 = -11 + 4 = -7
Check your answer:
The sum of the integers we found is: -11 + (-9) + (-7) = -27
We wanted it equal to five less than twice the smallest integer. The smallest integer is -11, twice the smallest integer is -22, less 5 is -27.

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