The sum of three consecutive integers is 13 greater than
twice the smallest of the three integers. Find the numbers. ( Since consecutive integers differ by 1 (1,2,3), we can
represent them as follows: let n represent the smallest of the three consecutive
integers; then n + 1 represents the second largest and , n + 2 represents the
largest . We can now create the following equation: n + (n + 1) + ( n + 2)
= 2n + 13 Solve this equation n + (n + 1) + ( n + 2)
= 2n + 13 n + n + 1 + n +
2 = 2n + 13 Remove
parentheses 3n + 3 = 2n + 13
Add the common terms 3n + 3 = 2n + 13
1n + 3 = 0n + 13 n + 3 = 13
n + 0 = 10 n = 10 Now we substitute the answer for n into the original equation to see if it works. n + (n + 1) + ( n + 2) = 2n + 13 10 + (10 + 1) + (10 + 2) = 2(10) + 13 10 + 11 + 12 = 20 + 13 33 = 33 We must also see if the solution works in the original problem statement. The sum of three consecutive integers is 13 greater than
twice the smallest of the three integers. The sum of three consecutive integers (10,11,12 are
consecutive integers) is 13 greater than twice the smallest of the three
integers (2 X 10). The sum of three consecutive integers (10 + 11 + 12 =
33) is 13 greater than twice the smallest of the three integers (2 X 10 =
20). The sum of three consecutive integers (33) is 13 greater
than twice the smallest of the three integers (20). 33 = 13 + 20 33 = 33 We have just solved a first degree equation. ( The source for the problem statement is: “Intermediate Algebra For College Students” by Jerome E. Kaufman |

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