algebra

posted by Bri

The value of a certain two-digit number is eight times he sum of its digits. if the digits of the number are reversed the result is 45 less than the original number. find the original number.

**please show both equations and work... thanks =)

  1. Reiny

    let the unit digit be x
    let the tens digit be y

    then 10y + x = 8(x+y)
    10y + x = 8x + 8y
    2y = 7x
    y = 7x/2

    the number reversed would be 10x+y

    then 10y+x - (10x+y) = 45
    9y -9x = 45
    -x+y=5
    sub in the 1st equation
    -x + 7x/2 = 5
    -2x + 7x = 10
    x = 2
    then y = 7

    original number is 72

  2. Anonymous

    when a number is squared ,the result is five times the original number find the number

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