The sum of three consecutive integers is 13 greater thantwice the smallest of the three integers. Find the numbers. ( Since consecutive integers differ by 1 (1,2,3), we canrepresent them as follows: let n represent the smallest of the three consecutiveintegers; then n + 1 represents the second largest and , n + 2 represents thelargest . 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 Removeparentheses 3n + 3 = 2n + 13Add the common terms 3n + 3 = 2n + 13
1n + 3 = 0n + 13 n + 3 = 13
n + 0 = 10 n = 10 Now we substitute the answerfor 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 thantwice the smallest of the three integers. The sum of three consecutive integers (10,11,12 areconsecutive integers) is 13 greater than twice the smallest of the threeintegers (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 greaterthan 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 AlgebraFor College Students by Jerome E. Kaufman |

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