Mathematics

Question

Find 4 consecutive odd integers , such that 5 times the second is equal to one more than eight times the smalller

2 Answer

  • Answer:

    The numbers are:

    3, 5, 7 and 9

    Step-by-step explanation:

    a = first odd number or "the smaller"

    a+2 = second consecutive odd number

    a+4 = third consecutive odd number

    a+6 = fourth consecutive odd number

    Then:

    5(a+2) = 1 + 8(a)

    (5*a + 5*2) = 1 + 8a

    5a + 10 = 1 + 8a

    10 - 1 = 8a - 5a

    9 = 3a

    9/3 = a

    a = 3

    a+2 = 5

    a+4 = 7

    a+6 = 9

    Check:

    5(5) = 1 + 8(3)

    25 = 1 + 24

  • Answer:

    • The integers are 3, 5, 7 an d 9

    -------------------------------------------------

    The difference between consecutive integers is 2.

    Let the smallest be n, then we have the following equation:

    • 5(n + 2) = 8n +1

    Solve it for n:

    • 5n + 10 = 8n + 1
    • 8n - 5n = 10 - 1
    • 3n = 9
    • n = 3

    The other integers are 5, 7, 9.

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