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Math Help - solving with 3 variables?!

  1. #1
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    solving with 3 variables?!

    Word problem: solving with 3 variables:

    Lila has three investments totaling $100,000. These investments earn interest at 7%, 9%, and 11% respectively. Lila's total income from these investments is $9800. The income from the 11% investment exceeds the total income from the other two investments by $1200. Find how much Lila has invested at 11%.

    Tasty Bakery sells three kinds of muffins: chocolate chip muffins at 40 cents each, oatmeal muffins at 45 cents each, and cranberry muffins at 50 cents each. Charles buys some of each kind and chooses twice as many cranberry muffins as chocolate chip muffins. If he spends $6.45 on 14 muffins, how many chocolate chip muffins did he buy?

    Solve the following system. What is the value of y?
    -5x - z = 2
    x + 4y - 2z = 1
    2x - 4y - z = -13

    Solve the following system. What is the value of y?
    -5x + 2z = 18
    x + 4y + z = -25
    -3x + 5y - z = -12

    Solve the following system. What is the value of y?
    -5x - z = 2
    x + 4y - 2z = 1
    2x - 4y - z = -13

    Please help! I am so lost..
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  2. #2
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    Quote Originally Posted by xbchgrl04x View Post
    Word problem: solving with 3 variables:

    Lila has three investments totaling $100,000. These investments earn interest at 7%, 9%, and 11% respectively. Lila's total income from these investments is $9800. The income from the 11% investment exceeds the total income from the other two investments by $1200. Find how much Lila has invested at 11%.

    ...
    Let a, b and c denote the amounts of the three investments. Then you know:

    \left|\begin{array}{rcl}a+b+c&=&100,000 \\ 0.07a+0.09b+0.11c&=&9800 \\ 0.07a+0.09b+1200&=&0.11c\end{array}\right.

    Choose the method to solve this system which you know best. You should get:
    (a; b; c) = (10,000 ; 40,000 ; 50,000)
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  3. #3
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    Quote Originally Posted by xbchgrl04x View Post
    ...
    Tasty Bakery sells three kinds of muffins: chocolate chip muffins at 40 cents each, oatmeal muffins at 45 cents each, and cranberry muffins at 50 cents each. Charles buys some of each kind and chooses twice as many cranberry muffins as chocolate chip muffins. If he spends $6.45 on 14 muffins, how many chocolate chip muffins did he buy?

    ...
    Let c denote the numbers of chocolate muffins, o the numbers of oatmeal muffins and b the numbers of cranberry muffins.

    Then you know:

    \left|\begin{array}{rcl}c+o+b&=&14 \\ 0.4c+0.45o+0.5b&=&6.45\\ b = 2c\end{array}\right.

    Choose the method to solve this system which you know best. You should get:
    (c; o; b) = ( 3; 5; 6)
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  4. #4
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    Quote Originally Posted by xbchgrl04x View Post
    ...

    Solve the following system. What is the value of y?
    -5x - z = 2
    x + 4y - 2z = 1
    2x - 4y - z = -13

    Solve the following system. What is the value of y?
    -5x + 2z = 18
    x + 4y + z = -25
    -3x + 5y - z = -12

    Solve the following system. What is the value of y?
    -5x - z = 2
    x + 4y - 2z = 1
    2x - 4y - z = -13

    Please help! I am so lost..
    These 2 (two) systems can be solved using one of the standard methods. The results are triples of integer numbers. Nothing very tricky!

    So I need to know where you have difficulties to do them. Otherwise I can't help you.
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  5. #5
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    Quote Originally Posted by earboth View Post
    These 2 (two) systems can be solved using one of the standard methods. The results are triples of integer numbers. Nothing very tricky!

    So I need to know where you have difficulties to do them. Otherwise I can't help you.
    Honestly, I get stuck from the begining...

    I have tried looking up examples online and this is what I get:

    -5x + 2z = 18
    x + 4y + z = -25
    -3x + 5y - z = -12

    -5x + 2z = 18
    x + 4y + z = -25
    -4x + 4y +32 = -7

    x + 4y + z = -25
    -3x + 5y - z = -12
    -2x + 9y = -37

    -4x + 4y + 3z = -7
    -2x + 9y = -37

    and then I am lost..
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  6. #6
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    Quote Originally Posted by xbchgrl04x View Post
    Honestly, I get stuck from the begining...

    ...
    Isn't the good ol' Google working anymore?

    Have a look here: System of linear equations - Wikipedia, the free encyclopedia
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    Quote Originally Posted by earboth View Post
    Isn't the good ol' Google working anymore?

    Have a look here: System of linear equations - Wikipedia, the free encyclopedia
    Okay, I tried them over again.. can you at least let me know if I got these examples right?

    Solve the following system. What is the value of y?
    4x - 5z = -33
    x + y + 2z = 11
    -x + 3y - z = 6
    y = 5

    Solve the following system. What is the value of y?
    -5x + 2z = 18
    x + 4y + z = -25
    -3x + 5y - z = -12
    y= -1

    Solve the following system. What is the value of y?
    -5x - z = 2
    x + 4y - 2z = 1
    2x - 4y - z = -13
    y= 3
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  8. #8
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    Quote Originally Posted by xbchgrl04x View Post
    Okay, I tried them over again.. can you at least let me know if I got these examples right?

    ...
    All three examples have triples of integer numbers as solution.

    My results differ from your's in all three cases.

    So I suggest that you do one example completely (not only publishing one value) so I can see where you need some additional assistance.
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  9. #9
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    Quote Originally Posted by xbchgrl04x View Post
    Solve the following system. What is the value of y?
    -5x - z = 2
    x + 4y - 2z = 1
    2x - 4y - z = -13
    I'll show you this one -- you try the others.

    I'm going to change the system into "triangular form," which looks something like this:
    \begin{aligned}<br />
x&\; +& ?y&\; +& ?z&\; = ? \\<br />
&\; {\color{white}.}& y&\; +& ?z&\; = ? \\<br />
&\; {\color{white}.}& &\; {\color{white}.}& z&\; = ?<br />
\end{aligned}
    ...where "?" are constants, and any of the "+"s could be minus signs.

    (To be continued below)


    01
    Last edited by yeongil; July 10th 2009 at 04:14 PM. Reason: Made a mess of things while editing...
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  10. #10
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    We start with this:
    \begin{aligned}<br />
-5x&\; {\color{white}.}& {\color{white}.}&\; -& z&\; = 2 \\<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
2x&\; -& 4y&\; -& z&\; = -13<br />
\end{aligned}

    Switch the position of Eq1 & Eq2:
    \begin{aligned}<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
-5x&\; {\color{white}.}& {\color{white}.}&\; -& z&\; = 2 \\<br />
2x&\; -& 4y&\; -& z&\; = -13<br />
\end{aligned}

    Multiply Eq1 by 5 and add the result to Eq2. Put the result in the Eq2 slot:
    \begin{aligned}<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
{\color{white}.}&\; {\color{white}.}& 20y&\; -& 11z&\; = 7 \\<br />
2x&\; -& 4y&\; -& z&\; = -13<br />
\end{aligned}

    Divide Eq2 by 20:
    \begin{aligned}<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
{\color{white}.}&\; {\color{white}.}& y&\; -& 0.55z&\; = 0.35 \\<br />
2x&\; -& 4y&\; -& z&\; = -13<br />
\end{aligned}

    Multiply Eq1 by -2 and add the result to Eq3. Put the result in the Eq3 slot:
    \begin{aligned}<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
{\color{white}.}&\; {\color{white}.}& y&\; -& 0.55z&\; = 0.35 \\<br />
{\color{white}.}&\; {\color{white}.}& -12y&\; +& 3z&\; = -15<br />
\end{aligned}

    Multiply Eq2 by 12 and add the result to Eq3. Put the result in the Eq3 slot:
    \begin{aligned}<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
{\color{white}.}&\; {\color{white}.}& y&\; -& 0.55z&\; = 0.35 \\<br />
{\color{white}.}&\; {\color{white}.}& {\color{white}.}&\; {\color{white}.}& -3.6z&\; = -10.8<br />
\end{aligned}

    Divide Eq3 by -3.6:
    \begin{aligned}<br />
x&\; +& 4y&\; -& 2z&\; = 1 \\<br />
{\color{white}.}&\; {\color{white}.}& y&\; -& 0.55z&\; = 0.35 \\<br />
{\color{white}.}&\; {\color{white}.}& {\color{white}.}&\; {\color{white}.}& z&\; = 3<br />
\end{aligned}

    You can see how easy it is from here to find x and y. I'll let you do that. (If you can see the white dots in the equations above, ignore them. I only put them there to line things up.)


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    Last edited by yeongil; July 10th 2009 at 04:14 PM. Reason: Made a mess of things while editing...
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  11. #11
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    When I think of systems of equations with three variables, I usually just use the Gaussian Method of Elimination...which is right here: Systems of Linear Equations: Gaussian Elimination
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