Question

A pizzeria sells three sizes of pizza: small, medium, and large. The pizzas sell for $6
, $10
, and $11
, respectively. One evening they sold 52
pizzas and received $419
. If they sold 7
more large than medium pizzas, how many of each size did they sell?

Use the Gaussian elimination method with back substitution to solve the given word problem.

Answer 1

Let's represent the number of small, medium, and large pizzas sold as x, y, and z respectively.

From the given information, we can create the following equations:

Equation 1: x + y + z = 52 (since they sold a total of 52 pizzas)
Equation 2: 6x + 10y + 11z = 419 (since the total revenue was $419)
Equation 3: z = y + 7 (since they sold 7 more large pizzas than medium pizzas)

We can rewrite Equation 1 as:
x + y + z = 52
1x + 1y + 1z = 52 (this will be the first row of our augmented matrix)

And rewrite Equation 2 as:
6x + 10y + 11z = 419
6x + 10y + 11z = 419 (this will be the second row of our augmented matrix)

And rewrite Equation 3 as:
0x + -1y + 1z = 7 (this will be the third row of our augmented matrix)

Now, we can construct the augmented matrix:
1 1 1 | 52
6 10 11 | 419
0 -1 1 | 7

We will perform Gaussian elimination to reduce the matrix to row-echelon form by using row operations.

1. Multiply Row 1 by -6 and add it to Row 2, then multiply Row 1 by 0 and add it to Row 3:
-6(1 1 1 | 52) + (6 10 11 | 419) -> (0 4 5 | 311)
0(1 1 1 | 52) + (0 -1 1 | 7) -> (0 -1 1 | 7)

The augmented matrix becomes:
1 1 1 | 52
0 4 5 | 311
0 -1 1 | 7

2. Multiply Row 2 by 1/4:
(1 1 1 | 52)
(0 1.0 1.25 | 77.75)
(0 -1 1 | 7)

The augmented matrix becomes:
1 1 1 | 52
0 1.0 1.25 | 77.75
0 -1 1 | 7

3. Add Row 2 to Row 3:
(1 1 1 | 52)
(0 1.0 1.25 | 77.75)
(0 0 2.25 | 84.75)

The augmented matrix becomes:
1 1 1 | 52
0 1.0 1.25 | 77.75
0 0 2.25 | 84.75

4. Divide Row 3 by 2.25:
(1 1 1 | 52)
(0 1.0 1.25 | 77.75)
(0 0 1 | 37.67)

The augmented matrix becomes:
1 1 1 | 52
0 1.0 1.25 | 77.75
0 0 1 | 37.67

Now, we will use back substitution to solve for x, y, and z.

From the third row, we know that z = 37.67.

From the second row, we know that 1.0y + 1.25z = 77.75
Substituting the value of z, we get: 1.0y + 1.25 * 37.67 = 77.75
1.0y + 47.08 = 77.75
1.0y = 30.67
y = 30.67

From the first row, we know that x + 1.0y + 1.0z = 52
Substituting the values of y and z, we get: x + 30.67 + 37.67 = 52
x + 68.34 = 52
x = -16.34

Since we cannot have pizza quantities in negative numbers, we disregard the negative value for x.

Therefore, they sold 0 small pizzas, 30 medium pizzas, and 37 large pizzas.