Problem 53 - College Algebra

Question

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. At a competing cupcake store, \(\$ 4,520\) worth of cupcakes are sold daily. The chocolate cupcakes cost \(\$ 2.25\) and the red velvet cupcakes cost \(\$ 1.75 .\) If the total number of cupcakes sold per day is \(2,200,\) how many of each flavor are sold each day?

Step-by-Step Solution

Answer

1,340 chocolate and 860 red velvet cupcakes are sold each day.

Step 1: Understanding the Variables

Let \( x \) be the number of chocolate cupcakes sold, and \( y \) be the number of red velvet cupcakes sold. We need to find the values of \( x \) and \( y \).

Step 2: Setting Up the Equations

The first equation represents the total number of cupcakes sold: \( x + y = 2200 \).The second equation represents the total sales: \( 2.25x + 1.75y = 4520 \).

Step 3: Building the Augmented Matrix

Using the two equations, we create the augmented matrix:\[\begin{bmatrix}1 & 1 & | & 2200 \2.25 & 1.75 & | & 4520\end{bmatrix}\]

Step 4: Row Reduction

Perform row operations to reduce the matrix to row-echelon form. Start by eliminating \( 2.25 \) in the second row by substituting \( R_2 \to R_2 - 2.25R_1 \):\[\begin{bmatrix}1 & 1 & | & 2200 \0 & -0.5 & | & -430\end{bmatrix}\]

Step 5: Solving for 'y'

From the second row, \( -0.5y = -430 \). Solve for \( y \):\[ y = \frac{-430}{-0.5} = 860 \]

Step 6: Substituting to Find 'x'

Substitute \( y = 860 \) back into the first equation: \( x + 860 = 2200 \).Solve for \( x \): \( x = 2200 - 860 = 1340 \).

Step 7: Verification

Verify that the number of cupcakes satisfies the original sales equation:\( 2.25 \times 1340 + 1.75 \times 860 = 4515 + 1505 = 4520 \).The calculations confirm the solution is correct.

Key Concepts

System of Equations

A system of equations consists of multiple equations that share common variables. In this exercise, we are dealing with two such equations, representing a real-world situation involving the sale of cupcakes. In our scenario, we need to determine the quantity of chocolate cupcakes and red velvet cupcakes sold daily. This situation can be expressed mathematically as:

• Equation 1: Represents total cupcakes sold: \( x + y = 2200 \)
• Equation 2: Represents total sales value: \( 2.25x + 1.75y = 4520 \)

The goal is to find values for \( x \) (chocolate cupcakes) and \( y \) (red velvet cupcakes) that satisfy both equations simultaneously. This ensures that the number of cupcakes and the sales match the given totals, effectively solving the system of equations.

Row Reduction

Row reduction, also known as Gaussian elimination, is a method for solving systems of linear equations by transforming a matrix into a simpler form. Specifically, a matrix can be row-reduced to row-echelon form, where the leading entry of each row is 1, and these leading 1s are strictly to the right of those in the previous row. With this exercise, once we have set up the augmented matrix, row reduction helps us simplify the system to efficiently solve for the unknowns. To achieve this, the matrix is modified using basic row operations, such as:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a multiple of one row from another row

These operations, as seen in the solution, help isolate one variable, making it straightforward to solve the system.

Matrix Row Operations

Matrix row operations are pivotal for manipulating matrices into a form that is easier to work with. In the context of an augmented matrix derived from a system of equations, these operations allow us to simplify and solve for unknown variables effectively. In the provided solution, the specific operation used is removing the coefficient in the second row by adjusting it with a multiple of the first row:1. Subtracting \( 2.25 \) times the first row from the second row to eliminate \( 2.25 \), creating a zero below the leading 1 in the first row.This operation:

• Makes the matrix easier to understand by focusing on one variable at a time.
• Lays the groundwork for back-substitution, a method for solving the simplified equations.

The focus of matrix row operations is always to make one variable stand out per row, gradually leading to a solution for all variables.

Solving Linear Equations

Solving linear equations is the process of finding values for variables that make all equations true. In this situation, once the system is expressed in an augmented matrix and simplified using row reduction, the next steps are straightforward arithmetic solutions. Following row reduction:1. The simplified matrix presents a clear path: \( -0.5y = -430 \).2. Solve for \( y \): multiply both sides by \( -1/0.5 \) to isolate \( y \).3. Using \( y = 860 \), substitute back into the original total equation \( x + 860 = 2200 \) to determine \( x \).4. Finally, verify by plugging these values back into both original equations to ensure both are satisfied, confirming the accuracy of the solution.This structured approach helps in breaking down complex problems into manageable steps, simplifying the task of finding solutions efficiently.