Find three consecutive even integers such that four
times the first minus the third is 6 more than twice the second. ( Since even consecutive integers differ by 2 (2,4,6), 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: 4(n) - (n + 4) =
2 (n + 2) + 6 Solve this equation 4(n) - (n +
4) =
2(n + 2) + 6 4n – n – 4 =
2n + 4 + 6 Remove parentheses 3n - 4 = 2n + 10 Add the
common terms 3n - 4 = 2n + 10
3n + 0 = 2n + 14 3n = 2n + 14
1n = 0n + 14 n = 14 Now we substitute the answer for n into the original equation to see if it works. 4(n) - (n + 4) =
2 (n + 2) + 6 4(14) - (14 + 4)
= 2 (14 + 2) + 6 4(14) - (18) = 2
(16) + 6 56 - 18 = 32 + 6 38 = 38 We must also see if the solution works in the original
problem statement. Find three consecutive even integers such that four
times the first minus the third is 6 more than twice the second. Find three consecutive even integers (14,16,18 are
consecutive even integers) such that four times the first (4 X 14) minus the
third (18) is 6 more than twice the second (2 X 16). Find three consecutive even integers such that four
times the first (56) minus the third (18) is 6 more than twice the second
(32). Find three consecutive even integers such that 56 minus
18 is 6 more than 32. 56 – 18 = 6 + 32 38 = 38 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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