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Math Help - Systems of Equations

  1. #1
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    Systems of Equations

    A chemist has three acid solutions at various concentrations. The first is 10% acid, the second is 20% acid, and the third is 40% acid. How many milliliters of each should he use to make 100 mL of 18% solution, if he has to use four times as much of the 10% solution as the 40% solution?

    I have no idea how to go about this. I'm sure I could solve the problem if I could just set it up! So can anyone help me set up a system of equations for this problem?

    EDIT: I've been contemplating this system:

    x + y + z = 100
    .1x + .2y + .4z = .18
    x = 4z

    But I don't know how I'd go about setting up a matrix for that...
    Last edited by blacksuzaku; June 3rd 2008 at 05:19 PM.
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  2. #2
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    Quote Originally Posted by blacksuzaku View Post
    A chemist has three acid solutions at various concentrations. The first is 10% acid, the second is 20% acid, and the third is 40% acid. How many milliliters of each should he use to make 100 mL of 18% solution, if he has to use four times as much of the 10% solution as the 40% solution?

    I have no idea how to go about this. I'm sure I could solve the problem if I could just set it up! So can anyone help me set up a system of equations for this problem?
    Let x=10% acid y=20% acid z=40% acid.

    We know that the total volume of the solution should be 100 mL so

    E_1: \\\ x+y+z=100 is the first equation

    For the next one the percents multiplied by the volume used need to add up so we get

    E_2: \\\ .10x+.20y+.40z=.18(100) \iff x+3y+4z=180

    Note: I multiplied the above equation by 10 to eliminate the decimals.

    For the last equation we know that he uses 4 times as much of 10% as 40%

    E_3: \\\ x=4z note that the equation is not 4x=z this is a common error.

    We now have a system of three equations with three unknowns.

    Good luck.
    Last edited by TheEmptySet; June 3rd 2008 at 05:52 PM.
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  3. #3
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    But I don't know how I'd go about setting up a matrix for that...
    \begin{bmatrix} 1 && 1 && 1 && 100 \\ 1 && 2 && 4 && 180 \\ 1 && 0 && -4 && 0 \end{bmatrix}
    Last edited by TheEmptySet; June 3rd 2008 at 05:52 PM.
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  4. #4
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    If you don't mind me asking, how does E2 become x + 3y + 4z = 180? Wouldn't it be x + 2y + 4z = 180?

    The second solution is 20% acid, not 30%. ^_^;;
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  5. #5
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    Quote Originally Posted by blacksuzaku View Post
    If you don't mind me asking, how does E2 become x + 3y + 4z = 180? Wouldn't it be x + 2y + 4z = 180?

    The second solution is 20% acid, not 30%. ^_^;;
    Yes you are correct. I will fix my above post.
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  6. #6
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    Aww, don't cry. Your answers are still extremely helpful and I thank you for life for helping me. After that, it was a simple matter to reduce the matrix to find a solution of x=50, y=40, and z=10. ^_^

    You have saved me from getting gray hairs at the tender age of 22.
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