Find three consecutive integers whose sume is 42. ( 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)= 42 Solve this equation n + (n + 1) + ( n + 2)= 42 n + n + 1 + n +2 = 42 Remove parentheses 3n + 3 = 42 Add the common terms 3n + 3 = 42
3n + 0 = 39 3n = 39 3n/3 = 39/3 Divide each side by 3 n = 13 Now we substitute the answerfor n into the original equation to see if it works. n + (n + 1) + ( n + 2)= 42 13 + (13 + 1) + (13 + 2) = 42 13 + 14 + 15 = 42 42 = 42 We must also see if the solution works in theoriginal problem statement. Find three consecutive integers whose sume is 42. Find three consecutive integers (13,14,15 areconsecutive integers) whose sume is 42. Find three consecutive integers (13 + 14 + 15 = 42)whose sume is 42. 42 = 42 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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