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Math Help - Systems of Equations Applications

  1. #1
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    Systems of Equations Applications

    Hello All! I have two word problems and I have no idea how to solve. Would be helpful if you could show steps and explain.

    1. Three cans of tuna fish and four cans of corned beef costs $12.50. However, six cans of tuna fish and three cans of corned beef cot $15. How much does each type of can cost individually?
    3.A school raised funds for its sports teams by selling tickets for a play. A ticket cost $5.75 for adults and $2.25 for students. If there were five times as many adults at the play than students and ticket sales raised $2480, how many adult and student tickers were sold?
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  2. #2
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    Quote Originally Posted by simoonlu View Post
    Hello All! I have two word problems and I have no idea how to solve. Would be helpful if you could show steps and explain.

    1. Three cans of tuna fish and four cans of corned beef costs $12.50. However, six cans of tuna fish and three cans of corned beef cot $15. How much does each type of can cost individually?
    3.A school raised funds for its sports teams by selling tickets for a play. A ticket cost $5.75 for adults and $2.25 for students. If there were five times as many adults at the play than students and ticket sales raised $2480, how many adult and student tickers were sold?
    Label the price of a can of tuna "x"
    and the price of a can of corned beef "y".

    Then we obtain a pair of simultaneous equations

    3x+4y=12.50

    6x+3y=15

    Now, eliminate one of x or y to find the other.
    You could continue as follows...

    2(3x+4y)=2(12.50)


    6x+8y=25

    6x+3y=15

    Subtracting the 2nd line from the first...

    6x+8y-(6x+3y)=25-15

    This allows us to find y, then use the value of y to find x.

    That's one way.
    How about the 2nd question ? Can you try it ?
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  3. #3
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    You can also use the coefficient matrix to solve the equations too.

    \displaystyle<br />
\begin{bmatrix}<br />
3 & 4 & 12.50\\ <br />
6 & 3 & 15<br />
\end{bmatrix}

    Using elementary row operations of first multiplying the top row by -2 and adding to the second and then dividing the second row by -5, you will obtain the y answer.

    \displaystyle<br />
\begin{bmatrix}<br />
3 & 4 & 12.50\\ <br />
6 & 3 & 15<br />
\end{bmatrix}\mbox{-2row1+row2} \Rightarrow\begin{bmatrix}<br />
3 & 4 & 12.50\\ <br />
0 & -5 & -10<br />
\end{bmatrix}

    \displaystyle<br />
\begin{bmatrix}<br />
3 & 4 & 12.50\\ <br />
0 & -5 & -10<br />
\end{bmatrix}\mbox{-1/5 row2}\Rightarrow \begin{bmatrix}<br />
3 & 4 & 12.50\\ <br />
0 & 1 & 2<br />
\end{bmatrix}

    Now, plug in y=2 into one of the equations and solve for x.
    Last edited by dwsmith; January 4th 2011 at 03:19 PM. Reason: Forgot the (-) sign
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