Find three consecutive odd integers such that threetimes the second minus the third is 11 more than the first. ( Since odd consecutive integers differ by 2 (1,3,5), wecan represent them as follows: let n represent the smallest of the three consecutiveintegers; then n + 2 represents the second largest and, n + 4 represents thelargest . We can now create the following equation: 3(n + 2) - (n + 4)= n + 11 Solve this equation 3(n + 2) - (n + 4)= n + 11 3n + 6n4= n + 11 Remove parentheses 2n + 2= n + 11 Add the common terms 2n + 2= n + 11
2n + 0 = n+ 9 2n = n + 9
1n = 0n + 9 n = 9 Now we substitute the answerfor n into the original equation to see if it works. 3(n + 2) - (n + 4)= n + 11 3(9 + 2) - (9 + 4)= 9 + 11 3(11) - (13)= 20 33 - 13 = 20 20 = 20 We must also see if the solution works in the originalproblem statement. Find three consecutive odd integers such that threetimes the second minus the third is 11 more than the first. Find three consecutive odd integers (9,11,13 areconsecutive odd integers) such that three times the second (3 X 11) minus thethird (13) is 11 more than the first (9). Find three consecutive odd integers such that threetimes the second (33) minus the third (13) is 11 more than the first. (9). Find three consecutive odd integers such that 33 minus13 is 11 more than 9. 33 - 13 = 11 + 9 20 = 20 We have just solved a first degree equation. ( The source for the problem statement is: Intermediate AlgebraFor College Students by Jerome E. Kaufman |

CloseWindow