Problem 53 - College Algebra with Corequisite Support

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

1260 chocolate and 940 red velvet cupcakes are sold daily.

Step 1: Define Variables

Let \( x \) be the number of chocolate cupcakes sold, and \( y \) be the number of red velvet cupcakes sold each day.

Step 2: Set Up Equations

Based on the problem, we have two equations: one for the total number of cupcakes and one for the total sales. 1. \( x + y = 2,200 \) (total cupcakes)2. \( 2.25x + 1.75y = 4,520 \) (total sales)

Step 3: Create Augmented Matrix

Write the system of equations as an augmented matrix: \[\begin{bmatrix}1 & 1 & | & 2200\2.25 & 1.75 & | & 4520\end{bmatrix}\]

Step 4: Use Row Operations to Simplify

Use row operations to simplify the augmented matrix and solve for \( x \) and \( y \). 1. Subtract \( 2.25 \times \text{(row 1)} \) from row 2 to eliminate \( x \) in the second row:\[\begin{bmatrix}1 & 1 & | & 2200\0 & -0.5 & | & 470\end{bmatrix}\]

Step 5: Solve for y

From the simplified matrix \( 0y - 0.5y = 470 \), we find:\(-0.5y = 470\)Solving this gives \( y = 940 \).

Step 6: Solve for x

Substitute \( y = 940 \) back into the first equation \( x + y = 2200 \):\( x + 940 = 2200 \)Solving this gives \( x = 1260 \).

Step 7: Verify Solution

Substitute \( x = 1260 \) and \( y = 940 \) into the sales equation to verify:\( 2.25 \times 1260 + 1.75 \times 940 = 2835 + 1645 = 4520 \)The solutions satisfy both equations, confirming they are correct.

Key Concepts

System of Equations

A "system of equations" is a set of equations with multiple variables that we need to solve at the same time. In this cupcake problem, we're given two equations. One represents the total count of cupcakes sold, and the other represents the total sales in dollars. These systems are crucial in finding specific unknowns — in our case, the different types of cupcakes sold. By solving such systems, we essentially find an intersection point where both equations are true. This is like finding a common solution that fits both conditions set by the equations. This aligns well with real-life scenarios where multiple conditions or rules need to be satisfied simultaneously.

Row Operations

"Row operations" are methods used to transform an augmented matrix when solving systems of equations. They involve three basic types: swapping two rows, multiplying a row by a non-zero constant, and adding or subtracting a multiple of one row from another. These operations help in simplifying matrices to find solutions. In our exercise, we use row operations to eliminate one variable at a time, turning a complex system into a more straightforward one. These operations are crucial as they preserve the solutions to the system, ensuring we won't lose or modify the real answer inadvertently.

Linear Algebra

"Linear algebra" is a branch of mathematics focusing on vectors, vector spaces, and linear equations. It's a powerful tool used to solve systems of equations efficiently. In the context of our problem, it helps us translate our system of equations into an **augmented matrix**. This visual approach not only simplifies computations but also provides a clear path to use mathematical operations for finding unknowns. Understanding the foundational principles of linear algebra is invaluable, as it underlies many real-world applications, including computer graphics, engineering, and economics.

Problem-Solving

"Problem-solving" in linear algebra involves breaking down a complex situation into manageable parts and systematically applying mathematical techniques to find solutions. For our cupcake store problem, several steps guide you: defining variables, setting up equations, creating an augmented matrix, simplifying using row operations, and finally solving for the variables. Tackling problems in stages helps ensure that no step is overlooked, improving accuracy and comprehension. This method is beneficial not only in math but in real-life decision-making processes as well, where logical step-by-step approaches yield the best outcomes.