college problem solving

a certain two-digit number is 6 less than the sum of its tens digit and 7 times its units digit. if the digits are reversed the number is increase by 27. fin the original two-digit numbers

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asked by luckybee
  1. let the tens digit be x and the unit digit be y

    then the number is 10x + y

    10x+y = (x+7y) - 6
    9x - 6y = -6
    3x - 2y = -2 , #1

    the number reversed would be 10y + x

    so 10y+x = 10x+y + 27
    9x - 9y = -27
    3x - 3y = -9 , #2

    #1 - #2
    y = 7
    then from #1, x = 4

    the number is 47

    check:
    reverse it --- 74
    is 74 -47 = 27 , so far so good

    what is the sum of its tens digit and 7 times its unit digit
    4 + 7(7) = 53
    is 53 greater than 47 by 6 ? YEAHHHH

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    posted by Reiny

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