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.
, $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.
Answers
GPT 3.5
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.
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.