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Math Help - Worded problem.

  1. #1
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    Worded problem.

    Out of school for 35 years and never taking algebra makes my kids homework a real challange, especially when my daughter is in honors math and looks to me for help. She works very hard to stay in class and I guess I have to keep up whith the times so I can be there for her and my son right behind her in school. After beating my head against the wall all week on this one problem I came acrossed this forum with a question. I can figure the answer the hard way but I am very curious about the formula. Question seems so simple:
    The class made $1220.00 on a school play they sold 295 tickets, adult tickets were $5 and children were $2 how many Adult tickets were sold.

    Thank You

    Jim
    Last edited by mr fantastic; January 2nd 2011 at 08:33 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by Sven22 View Post
    The class made $1220.00 on a school play they sold 295 tickets, adult tickets were $5 and children were $2 how many Adult tickets were sold.
    Let a = adult tickets sold; then children tickets sold = 295 - a ; OK?

    So 5a + 2(295 - a) = 1220 ; OK?

    Can you solve that for a?
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  3. #3
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    295=a+c

    5a+2c=1220

    \displaystyle \begin{bmatrix}<br />
1 & 1 & 295\\ <br />
5 & 2 & 1220<br />
\end{bmatrix}\Rightarrow \begin{bmatrix}<br />
1 & 1 & 295\\ <br />
0 & -3 & -255<br />
\end{bmatrix}\Rightarrow\begin{bmatrix}<br />
1 & 1 & 295\\ <br />
0 & 1 & 85<br />
\end{bmatrix}

    c=85

    I used elementary row operations to solve the matrices: addition/subtraction of rows, multiplying by a scalar, and both operations combined.

    The first was multiple row 1 by 5 and subtract it from the second.

    Then, I divide row 2 by -3.
    Last edited by dwsmith; January 2nd 2011 at 05:26 PM.
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  4. #4
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    This is a simultaneous equations question.

    Let \displaystyle a represent the number of adult tickets and \displaystyle c represent the number of children's tickets.

    There were \displaystyle 295 tickets sold, so \displaystyle a+c = 295.

    They made \displaystyle \$5 per adult ticket and \displaystyle \$2 per children's ticket, getting a total of \displaystyle \$1220. So \displaystyle 5a + 2c = 1220.


    Now you have a system of equations:

    \displaystyle a+c = 295
    \displaystyle 5a+2c=1220.

    Solve them simultaneously for \displaystyle a and \displaystyle c.
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  5. #5
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    Another way to do is to solve as a simple system:

    Let 'a' be the number adult tickets sold, and 'c' be the number of children tickets.

    a + c = 295
    5a + 2c = 1220

    Solve for 'c' in the first:
    c = 295 - a

    Sub 'c = 295 - a' into the second:
    5a + 2(295 - a) = 1220
    5a + 590 - 2a = 1220
    3a = 630
    a = 210

    Which is the number of adult tickets sold. The number of children can be found by subbing back into the first equation with a = 210.

    210 + c = 295
    c = 85

    Edit: Beaten!
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    so the algerbraic expression to solve this problem is 5a+2(295-a)=1220
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  7. #7
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    Quote Originally Posted by Sven22 View Post
    so the algerbraic expression to solve this problem is 5a+2(295-a)=1220
    Solve for a.

    Plug the value of a into the equation a+c=295 to obtain c.
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