ALgebra...plz help

If the tens digit of a two-digit number is subtracted from the units digit, the difference is 8. The number with the digits reversed is 10 more than nine times the units digit of the original number. Find the number.

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  1. Let the two digit number be 10A + B.

    Then, from "If the tens digit of a two-digit number is subtracted from the units digit, the difference is 8.", or
    B - A = 8.

    From "The number with the digits reversed is 10 more than nine times the units digit of the original number.", 10B + A = 9B + 10.

    Substituting A = B - 8 into 10B + A = 9B + 10 yields 10B + B - 8 = 9B + 10 or 11B - 8 = 9B + 10 or 2B = 18 making B = 9.

    From B - A = 8, 9 - A = 8 making A = 1 or the number = 19.

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  2. too easy

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  3. Let unit digit = x
    Then tens did it = 8+x
    Original no.is 10(8+x)+x
    80+10x+x=80+11x
    On reversing : 10x+8+x =11x+8
    11x+8=10+9x
    11x-9x=10-8
    2x=2
    X=1
    No.is 80+11x=80+11=91

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