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

Question

asked Nov 25, 2019 235k views

1 Answer

Lee Schmidt · Answer 1 · 2019-12-01T15:51:51+0000

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