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Math Help - coin-word problems in algebra

  1. #1
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    coin-word problems in algebra

    Hi everyone,

    I am doing lots of word problems and so far I was doing pretty good. For some reason I can't figure this one out. I am hoping someone is able to help me.

    In a two-digit number , the ten's digit is 3 more than the one's digit. If the digits are reversed, the difference between the two numbers is 27. Find the number.

    ten's digit= x+3 one's digit = x
    10(x+3) + x
    reversed would be 10x + x+3 -27 = 10(x+3)+x ??? I don't get it.
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  2. #2
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    Quote Originally Posted by Christine Magruder View Post
    Hi everyone,

    I am doing lots of word problems and so far I was doing pretty good. For some reason I can't figure this one out. I am hoping someone is able to help me.

    In a two-digit number , the ten's digit is 3 more than the one's digit. If the digits are reversed, the difference between the two numbers is 27. Find the number.

    ten's digit= x+3 one's digit = x
    10(x+3) + x
    reversed would be 10x + x+3 -27 = 10(x+3)+x ??? I don't get it.
    number: 10x + 30 + x
    reversed : 10x + x + 3

    10x+ 30 + x - 10x - x -3 = 27
    0x = 0

    in this case, x would be all the positive real values but check the equation by substituting values.
    x = 1 (possible)
    x = 2 (possible)
    x = 3 (possible)
    x = 4 (possible)
    x = 5 (possible)
    x = 6 (possible)
    x = 7 (rejected since it gives a three digit number)

    so x = + {1,2,3,4,5,6}
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