Find three consecutive integers whose sume is 42. ( 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)
= 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 answer for 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 the
original problem statement. Find three consecutive integers whose sume is 42. Find three consecutive integers (13,14,15 are
consecutive 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 Algebra For College Students” by Jerome E. Kaufman |

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