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March 26, 2017

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Find 4 consecutive odd integers such that 5 times the sum of the first two was 10 less than 7 times the sum of the second and fourth. What are the four integers?

When I worked the problem, I got -13 as N and then when I finished it, it didn't seem correct. Can anyone help me?

  • Algebra II - ,

    let the smallest of the odd integers be x
    then the next 3 consecutive integers are
    x+2,x+4, and x+6

    5(x + x+2) = 7(x+2 + x+6) - 10
    10x + 10 = 14x + 46
    -4x = 36
    x = -9

    so the 4 integers are -9, -7, -5, and -3

    check:
    5times(sum of first two) = 5(-9 -7) = -80
    7times(sum of 2nd and 4th) = 7(-7 -3) = -70

    Is -80 ten less than -70? YES

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