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Thread: Finding the two numbers... (check my work)

  1. #1
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    Finding the two numbers... (check my work)

    Hey, everyone!

    I'm having a little trouble with this one... I'm pretty sure I'm missing something, though. Here's the problem:

    "The product of two numbers is 407. One of the numbers is three times the other number, augmented by four. Find the sum of the two numbers."

    So, I chose $\displaystyle x$ and $\displaystyle y$ as my numbers and this is what I got, based on the description:

    $\displaystyle x$
    $\displaystyle y=3x+4$

    When multiplied, you get 407

    $\displaystyle xy=407$

    So, I substitute and simplify:

    $\displaystyle x(3x+4)=407$
    $\displaystyle 3x^2+4x=407$
    $\displaystyle 3x^2+4x-407=0$
    $\displaystyle (3x+37)(x-11)=0$

    $\displaystyle x=-\frac{3}{37}, 11$

    I'm assuming the solution I should choose is the 11, right?

    Now, substituting for $\displaystyle x=11$

    $\displaystyle y=3x+4$
    $\displaystyle y=3(11)+4$
    $\displaystyle y=37$

    and now I add:

    $\displaystyle 37 + 11 = 48$

    Thoughts?

    Thanks in advance!
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  2. #2
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    Re: Finding the two numbers... (check my work)

    You will have to show BOTH possible sets of solutions.
    Thanks from Tall Jessica
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  3. #3
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    Re: Finding the two numbers... (check my work)

    Quote Originally Posted by Prove It View Post
    You will have to show BOTH possible sets of solutions.
    So, it's just a matter of plugging in the other solution, too? Is what I've done correct, then?

    Thank you so much!!
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