Problem 104 - College Algebra with Modeling and Visualization

Question

Mixing Candy Two types of candy sell for \(\$ 2.50\) per pound and \(\$ 4.00\) per pound. A store clerk is trying to make a 5 -pound mixture worth \(\$ 17.60 .\) How much of each type of candy should be included in the mixture?

Step-by-Step Solution

Answer

1.6 pounds of \$2.50/lb candy and 3.4 pounds of \$4.00/lb candy.

Step 1: Define Variables

Let \(x\) be the weight in pounds of the \\(2.50/lb candy, and \(y\) be the weight in pounds of the \\)4.00/lb candy.

Step 2: Set Up Equations Based on Total Weight

We know that the total weight of the candy mixture is 5 pounds. Thus, \(x + y = 5\).

Step 3: Set Up Equations Based on Total Price

The total price of the mixture is \$17.60. Therefore, the equation for the total cost of the candies is \(2.50x + 4.00y = 17.60\).

Step 4: Solve the System of Equations

We have two equations: \(x + y = 5\) and \(2.50x + 4.00y = 17.60\). Solve the first equation for \(x\): \(x = 5 - y\). Substitute into the second equation: \(2.50(5 - y) + 4.00y = 17.60\). Simplify: \(12.5 - 2.50y + 4.00y = 17.60\). Combine like terms: \(1.50y = 5.10\). Solve for \(y\): \(y = \frac{5.10}{1.50} \approx 3.4\).

Step 5: Calculate the Amount of First Type of Candy

Substitute \(y = 3.4\) back into \(x = 5 - y\): \(x = 5 - 3.4 = 1.6\).

Step 6: Verify the Solution

Check if the calculated weights satisfy the conditions: \(2.50 \times 1.6 + 4.00 \times 3.4 = 4.00 + 13.60 = 17.60\). The solution satisfies the equation, confirming the result is correct.

Key Concepts

Variable Definition

When tackling problems that involve a system of linear equations, the first important step is to define the variables clearly. This means assigning symbols to unknown quantities in the problem. In our exercise, the goal is to find out how much of each type of candy we need to create a 5-pound mixture worth \(17.60. For this, we define:

• \(x\): the weight in pounds of the candy costing \)2.50 per pound.
• \(y\): the weight in pounds of the candy costing $4.00 per pound.

By defining these variables, we can set up equations that will help us solve the problem effectively. Defining variables precisely is crucial, as it sets the groundwork for the solution.

Equation Setup

The next step in solving a system of linear equations is setting up the correct equations based on the information provided. We need to reflect two requirements: the total weight of the mixture and the total price.For total weight, we know:

• \(x + y = 5\) , because the mixture weighs a total of 5 pounds.

Then, we address the total cost:

• \(2.50x + 4.00y = 17.60\), stating that the blend should cost $17.60.

These equations form the system that we need to solve. Understand that each equation corresponds to a specific condition or requirement from the problem.

Problem Solving Steps

With the equations ready, solving them involves step-by-step algebraic manipulations. We want to find values for \(x\) and \(y\) that satisfy both equations.First, solve one of the equations for one variable. Here, we take the weight equation:

• \(x = 5 - y\)

Substituting this into the cost equation, we replace \(x\) with \(5-y\) in\(2.50x + 4.00y = 17.60\). This yields:

• \(2.50(5-y) + 4.00y = 17.60\)
• Expand and simplify: \(12.5 - 2.50y + 4.00y = 17.60\)
• Combining like terms gives \(1.50y = 5.10\)
• Solve for \(y\): \(y = \frac{5.10}{1.50} \approx 3.4\)

Once \(y\) is found, substitute back to compute \(x\):

• \(x = 5 - 3.4 = 1.6\)

Verification of Solution

After arriving at a solution, it is essential to verify that it satisfies the original problem conditions. This ensures that our solution is both correct and reliable.Substitute the calculated values of \(x = 1.6\) and \(y = 3.4\) back into the equations:

• Total weight check: \(x + y = 1.6 + 3.4 = 5\), which matches the given total weight of 5 pounds.
• Total cost check: Compute \(2.50 \times 1.6 + 4.00 \times 3.4\), which equals \(4.00 + 13.60 = 17.60\), confirming the cost matches the desired $17.60.

Because both conditions are satisfied, our solution is valid. Verification not only confirms accuracy but also builds confidence in the problem-solving process.