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20 April, 06:12

Find three consecutive even integers, such that 9 times of first integer is 8 more than the sum of second and third integers

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  1. 20 April, 06:34
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    Let the middle number = X

    The other two numbers are even so the lowest number would be X - 2, and the highest number would be X + 2.

    9 times the first number (lowest number) would be 9 (x-2) = 9x-18

    The sum of the other two numbers becomes x + x+2 = 2x+2

    The first number is 8 more so now you have an equation of:

    9x-18 = 2x+2 + 8

    Combine like terms on the right side:

    9x-18 = 2x+10

    Add 18 to both sides:

    9x = 2x + 28

    Subtract 2x from each side:

    7x = 28

    Divide both sides by 7:

    X = 28 / 7 = 4

    The middle number is 4.

    The lowest number = X-2 = 4-2 = 2

    The highest number = X+2 = 4+2 = 6

    The 3 numbers are 2,4 & 6
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