To help passengers stranded by bad weather one winter, an airport bought cots, chairs, and tables totaling 35. They bought cots at $45 each, chairs for $15 each, and tables for $60 each. They spent a total of $1350 on the furniture. The number of chairs they purchased was 10 less than the number of cots. Set up a system and an augmented matrix to solve for the number of each type of furniture pieces the airport purchased.

[tex]\begin{bmatrix}{1} & {1} & {1} \\ {45} & {15} & {60} \\ {1} & {-1} & {0}\end{bmatrix}\begin{bmatrix}{x} \\ {y} \\ {z}\end{bmatrix}=\begin{bmatrix}{35} \\ {1350} \\ {10}\end{bmatrix}[/tex]

Explanation:

Let x, y and z represent the number of cots, chairs, and tables the airport bought.

Given;

The airport bought cots, chairs, and tables totaling 35. the total number of cots, chairs and tables bought is 35. So, we have;

[tex]x+y+z=35\text{ ---------1}[/tex]

If they bought cots at $45 each, chairs for $15 each, and tables for $60 each. And they spent a total of $1350 on the furniture.

Then the total cost is the sum of the cost of each type of furniture;

[tex]45x+15y+60z=1350\text{ ----------2}[/tex]

And also, The number of chairs they purchased was 10 less than the number of cots. So, we have;

[tex]\begin{gathered} y=x-10 \\ x-y=10\text{ --------3} \end{gathered}[/tex]

So we have the three equations needed.

We can combine the 3 equations to form a matrix.

Therefore, the system of matrix needed to solve the problem will be;

[tex]\begin{bmatrix}{1} & {1} & {1} \\ {45} & {15} & {60} \\ {1} & {-1} & {0}\end{bmatrix}\begin{bmatrix}{x} \\ {y} \\ {z}\end{bmatrix}=\begin{bmatrix}{35} \\ {1350} \\ {10}\end{bmatrix}[/tex]