##### Matrix word problem help?

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You sell tickets at school for fundraisers.You sold car wash tickets, silly string fight tickets and dance tickets, for a total of 380 tickets sold.The car wash tickets were \$5 each, the silly sting fight tickets were \$3 each and the dance tickets were \$10 each.If you sold twice as many silly string tickets as car wash tickets,and you have \$1460 total,how many dance tickets were sold?

1. Write the system of linear equations

2. Write the matrix

3. Write the solution set

Oct 2nd, 2015

Thank you for the opportunity to help you with your question! I'll be explaining how I got to the answer, but if you don't want to read all of the explanations, I'll just highlight my work.

We first need to figure out what the problem wants and what we information we can take from the problem:

From the first sentence, there are three possible variables dealing with the car wash tickets (x), silly string fight tickets (y), and the dance tickets (z). They also give a total amount of tickets, 380, meaning that 380 tickets is the sum of x, y, and z tickets sold. From this, we can make the equation:
x + y + z = 380 (where x, y, and z are variables representing the numbers of each ticket item sold)

The next sentence tells us the prices of each ticket and the total amount of money you got from selling: \$1460. Thus, we can make another equation in which the prices of each ticket multiplied to their respective ticket types. From the information, we get this equation
5x + 3x + 10z = 1460 (because car wash tickets are \$5 each, etc, etc)

Now that we have at least two equations, we can solve the system using an augmented matrix, where you have rows associated with each equation. (Please let me know if I should clarify that). There are many ways to do this, but I personally want to work with nicer looking matrices, so I simplified the equations a little bit. If you want the other ways to solve it, I have also worked those out and can show you.

Using the last portion of the question, we can see that the number of silly string tickets sold (y) is equal to twice the number of car wash tickets sold (2x), so you just plug 2x into both equations to get:

3x + z = 380 and 11 + 10x = 1460 (be careful to account for the other coefficients in this one!)

Now, our augmented matrix looks like this:
[    3    1     |     380
11   10   |    1460    ] (the | is purely for visual purposes; it divides the variable coefficients from the solutions)

Now that we have a matrix, we can use basic row operations to reduce it into reduced echelon form (I've done the work on paper, but it's a little much to put on here. I can show you if you'd like to check your work). We should get this matrix:

[ 1     0  |   2340/19       which can be rewritten as this:  [ 1  0    * [ x     =   [ 2340/19
0     1  |    200/19  ]                                                         0  1 ]     z ]              200/19 ]

Algebraically, it means x = 2340/19, z = 200/19.

To calculate y, we can use x since we know that y = 2x: y = 2*(2340/19) = 4680/19.

The solution set is:
x = 2340/19, y = 4680/19, z = 200/9 (or the decimal value rounded to the nearest whole)

and the number of dance tickets that were sold is:
z = 4680/19 ≈ 246 tickets

I hope this wasn't too long for you and was clear enough. Please let me know if you need any clarification. I'm always happy to answer your questions! Thank you!
Oct 3rd, 2015

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Oct 2nd, 2015
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Oct 2nd, 2015
Sep 23rd, 2017
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