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Question 839702: find three consecutive odd integers such that twice the sum of the first and third integers is 21 more than the second integer. define a variable, set up and equation, and solve it, and give the final answer clearly,
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! since the numbers are consecutive odd numbers, then they have to be separated from each other by 2.
let your first number be x.
the second number is then x + 2
the third number is then x + 4
you are given that 2 times the sum of the first and third numbers is 21 more than the second number.
translate this to pseudo algebra to get:
2 * (first number + third number) = second number + 21
replace first number with x.
replace second number with (x + 2)
replace third number with (x + 4)
your equation becomes:
2 * (x + x + 4) = x + 2 + 21
combine like terms to get:
2 * (2x + 4) = x + 23
simplify further by distributing the multiplication on the left side of the equation to get:
4x + 8 = x + 23
subtract x from both sides of the equation and subtract 8 from both sides of the equation to get:
4x - x = 23 - 8
simplify to get:
3x = 15
divide both sides of this equation by 3 to get:
x = 5
your numbers are 5, 7, 9
two times the sum of the first number and the third number is equal to 28.
the second number and 21 is equal to 28.
two times the sum of the first number and the third number is equal to the sum of the second number and 21 because 28 is equal to 28 so you're good and the numbers shown are your solution.
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