Creating System of Linear equations

Linear spaces and subspaces, linear transformations, bases, etc.
johnny23o
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a movie theater charges $9.00 for adults and$7.00 for children. On a day when 325 people purchased tickets, the total receipts were $2675. How many adults tickets were sold? How many children tickets were sold? Create a system of linear equations and solve it using matrices. I am really lost here please help Martingale Posts: 350 Joined: Mon Mar 30, 2009 1:30 pm Location: USA Contact: Re: Creating System of Linear equations johnny23o wrote:a movie theater charges$9.00 for adults and $7.00 for children. On a day when 325 people purchased tickets, the total receipts were$2675. How many adults tickets were sold? How many children tickets were sold? Create a system of linear equations and solve it using matrices.

Let Y=#children tickets

now make two liner equations

johnny23o
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Re: Creating System of Linear equations

Let x= $9.00 let y=$7.00

9x+7y-325z=2675

would this work?

Martingale
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Re: Creating System of Linear equations

johnny23o wrote:Let x= $9.00 let y=$7.00

9x+7y-325z=2675

would this work?

no...you are letting x and y equal costs...let them equal the number of adult and children respectively. you should get two equations not one.

stapel_eliz
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johnny23o wrote:Let x= $9.00 let y=$7.00

9x+7y-325z=2675

would this work?

To learn the basic method for setting up this sort of exercise, try here.

Jaroslaw Jackiw
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Joined: Sat Sep 11, 2010 2:44 am
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Re: Creating System of Linear equations

Your problem states that $325$ people attended the theater that day, grossing $\2675$, at $\9$ per ticket per adult ($x_1$)and $\7$ per ticket per child ($x_2$). This means that $\9$ times $x_1$, the number of adults, plus $\7$ times $x_2$, the number of children, produced $\2675$, and that the total number of adults, $x_1$, plus children, $x_2$, was $325$. This is your system of linear equations:

\begin{align}
9x_1 + 7x_2 &= 2675\\
x_1 + x_2 &= 325
\end{align}

You mentioned that you needed to use matrices to solve the problem. Then you will need to convert your system of linear equations to matrix form, as follows:

$\begin{equation}
\begin{bmatrix}
9 & 7\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
=
\begin{bmatrix}
2675\\
325
\end{bmatrix}
\end{equation}$

This will produce a matrix equation of the form $AX = B$ for:

\begin{align}
A &= \begin{bmatrix}
9 & 7\\
1 & 1
\end{bmatrix}\\
X &= \begin{bmatrix}
x_1\\
x_2
\end{bmatrix}\\
B &= \begin{bmatrix}
2675\\
325
\end{bmatrix}\\
I &= \begin{bmatrix}
1 &0\\
0 &1
\end{bmatrix}
\end{align}

such that:

\begin{align*}
AX &= B\\
A^{-1}AX &= A^{-1}B\\
IX &= A^{-1}B\\
X &= A^{-1}B
\end{align*}

Given $X$ and $B$, you need to find $A^{-1}$, the inverse matrix of $A$, if it exists, to solve the equation. Using elementary row operations on $A^*$, the augmented matrix of $A$, allows you to obtain $A^{-1}$:

\begin{align*}
&= \begin{bmatrix}
9 &7 &| &1 &0\\
1 &1 &| &0 &1
\begin{bmatrix}
1 &0 &| &\frac{1}{2} &-\frac{7}{2}\\
0 &1 &| &-\frac{1}{2} &\frac{9}{2}
\end{align*}

So that:

\begin{align*}
A^{-1} &=
\begin{bmatrix}
\frac{1}{2} &-\frac{7}{2}\\
-\frac{1}{2} &\frac{9}{2}
\end{bmatrix}\\

&= \frac{1}{2}
\begin{bmatrix}
1 &-7\\
-1 &9
\end{bmatrix}
\end{align*}

Since $X = A^{-1} B$ for:

\begin{align}
X &=
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}\\

A^{-1} &= \frac{1}{2}
\begin{bmatrix}
1 &-7\\
-1 &9
\end{bmatrix}\\

B &=
\begin{bmatrix}
2675\\
325
\end{bmatrix}
\end{align}

then:

\begin{align*}
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
&= \frac{1}{2}
\begin{bmatrix}
1 &-7\\
-1 &9
\end{bmatrix}
\begin{bmatrix}
2675\\
325
\end{bmatrix}\\
&= \frac{1}{2}
\begin{bmatrix}
2675 - 7 \cdot 325\\
9 \cdot 325 - 2675
\end{bmatrix}\\
&= \frac{1}{2}
\begin{bmatrix}
2675-2275\\
2925-2675
\end{bmatrix}\\
&= \phantom{\frac{1}{2}}
\begin{bmatrix}
200\\
125
\end{bmatrix}
\end{align*}

Therefore, $x_1=200$ adults and $x_2=125$ children attended the theater that day.