Find four consecutive integers whose sume is -118. ( Since consecutive integers differ by 1 (1,2,3), we canrepresent them as follows: let n represent the smallest of the four consecutiveintegers; then n + 1 represents the second largest, n + 2 represents the third largest, and n + 3 representsthe largest . We can now create the following equation: n + (n + 1) + (n + 2) + (n + 3) = -118 Solve this equation n + (n + 1) + (n + 2) + (n + 3) = -118 n + n + 1 + n +2 + n + 3 = -118Remove parentheses 4n + 6 = -118Add the common terms 4n + 6 = -118
4n + 0 = -124 4n= -124 4n/4= -124/4 Divide each side by 4 n =-31 Now we substitute the answerfor n into the original equation to see if it works. n + (n + 1) + (n + 2) + (n + 3) =-118 -31 + (-31 + 1) + (-31 + 2)(-31 + 3) = -118 -31 + -30 + -29 + -28 = -118 -118 = -118 We must also see if the solution works in theoriginal problem statement. Find four consecutive integers whose sume is 118. Find four consecutive integers (-31, -30, -29,-28 are consecutive integers) whose sume is -118. Find four consecutive integers (-31 + -30 + -29 + -28 = -118) whose sume is -118. -118 = -118 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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