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28 February, 19:52

find two consecutive odd intergers that twice the larger is fifteen more than three times the smaller

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Answers (2)
  1. 28 February, 20:50
    0
    11, 13

    Step-by-step explanation:

    Let n and n+2 be the consecutive odd integers

    2 (n+2) = 3n + 15

    2n + 4 = 3n + 15

    n = 11

    n + 2 = 13
  2. 28 February, 22:47
    0
    11, 13

    Step-by-step explanation:

    an odd number can be represented by 2n+1

    since they are consecutive, the larger odd number will be 2n + 1 + 2

    now,

    2 (2n+3) = 2n + 1 + 15

    solving this eqn, we get n = 5

    so the two numbers are (2n+1) = 11 and 11+2 = 13
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