## Creating System of Linear equations

Linear spaces and subspaces, linear transformations, bases, etc.
johnny23o
Posts: 4
Joined: Mon Aug 09, 2010 10:59 pm
Contact:

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: 333 Joined: Mon Mar 30, 2009 1:30 pm Location: USA Contact: ### Re: Creating System of Linear equations 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
Posts: 4
Joined: Mon Aug 09, 2010 10:59 pm
Contact:

### Re: Creating System of Linear equations

Let x= $9.00 let y=$7.00

9x+7y-325z=2675

would this work?

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
Contact:

### Re: Creating System of Linear equations

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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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
Posts: 1
Joined: Sat Sep 11, 2010 2:44 am
Contact:

### 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*} A^* &= [A|I] \quad \rightarrow{} \quad \cdots \quad \rightarrow{} \quad [I|A^{-1}] = A^*_t\\ &= \begin{bmatrix} 9 &7 &| &1 &0\\ 1 &1 &| &0 &1 \end{bmatrix} \quad \rightarrow{} \quad \cdots \quad \rightarrow{} \quad \begin{bmatrix} 1 &0 &| &\frac{1}{2} &-\frac{7}{2}\\ 0 &1 &| &-\frac{1}{2} &\frac{9}{2} \end{bmatrix} \quad = A^*_t \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.