Find three consecutive odd integers such that three
times the second minus the third is 11 more than the first. ( Since odd consecutive integers differ by 2 (1,3,5), we
can represent them as follows: let n represent the smallest of the three consecutive
integers; then n + 2 represents the second largest and, n + 4 represents the
largest . 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 + 6 – n – 4
= 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 answer for 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 original
problem statement. Find three consecutive odd integers such that three
times the second minus the third is 11 more than the first. Find three consecutive odd integers (9,11,13 are
consecutive odd integers) such that three times the second (3 X 11) minus the
third (13) is 11 more than the first (9). Find three consecutive odd integers such that three
times the second (33) minus the third (13) is 11 more than the first. (9). Find three consecutive odd integers such that 33 minus
13 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 Algebra For College Students” by Jerome E. Kaufman |

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