Find four consecutive odd integers such that 3 times the largest decreased by 19 is twice the second integer

n, n + 2, n + 4, n + 6 - four consecutive odd integers

3 times the largest decreased by 19 is twice the second integer

3(n + 6) - 19 = 2(n + 2)     use distributive property

(3)(n) + (3)(6) - 19 = (2)(n) + (2)(2)

3n + 18 - 19 = 2n + 4

3n - 1 = 2n + 4      add 1 to both sides

3n = 2n + 5      subtract 2n from both sides

n = 5

n + 2 = 5 + 2 = 7

n + 4 = 5 + 4 = 9

n + 6 = 5 + 6 = 11

x=19

y=14.5

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