Problem 33 - Algebra A Combined Function

Question

Wayne Osby blends coffee for a local coffee café. He needs to prepare 200 pounds of blended coffee beans selling for \(\$ 3.95\) per pound. He intends to do this by blending together a high-quality bean costing \(\$ 4.95\) per pound and a cheaper bean costing \(\$ 2.65\) per pound. To the nearest pound, find how much high-quality coffee bean and how much cheaper coffee bean he should blend.

Step-by-Step Solution

Answer

Use 113 pounds of high-quality beans and 87 pounds of cheaper beans.

Step 1: Set Up Variables for Beans

Let \( x \) represent the number of pounds of high-quality coffee beans that cost \\(4.95 per pound. Let \( y \) represent the number of pounds of cheaper coffee beans that cost \\)2.65 per pound. We need to find the values of \( x \) and \( y \).

Step 2: Establish the Total Weight Equation

Wayne needs a total of 200 pounds of blended coffee. Therefore, we can write the equation for the total weight as: \( x + y = 200. \)

Step 3: Establish the Total Cost Equation

The desired price of the blended coffee is \$3.95 per pound for 200 pounds. This gives a total cost of \( 200 \times 3.95 = 790 \) dollars. Hence, the equation for the total cost is: \( 4.95x + 2.65y = 790. \)

Step 4: Solve the System of Equations

We now have two equations: \( x + y = 200 \) and \( 4.95x + 2.65y = 790 \). We can use substitution or elimination to solve for \( x \) and \( y \). First, solve the first equation for \( y \): \( y = 200 - x \) and substitute into the second equation to solve for \( x \): \[4.95x + 2.65(200 - x) = 790\]Simplify and solve for \( x \):\[4.95x + 530 - 2.65x = 790\]\[2.3x = 260\]\[x = 113\]Now substitute \( x = 113 \) into \( y = 200 - x \) to find \( y \):\[y = 200 - 113 = 87\]

Step 5: Verify the Solution

Substitute \( x = 113 \) and \( y = 87 \) back into the equation \( 4.95x + 2.65y = 790 \) to ensure that the solution is correct:\[4.95(113) + 2.65(87) = 559.35 + 230.55 = 790\]Since both equations are satisfied, the solution is correct.

Key Concepts

Substitution Method

The substitution method is a commonly used technique to solve systems of linear equations. In this approach, we first solve one of the equations for one of the variables, and then substitute this expression into the other equation. This helps us eliminate one of the variables and solve for the other.

In Wayne Osby's coffee blending problem, we used the equation for the total weight, \(x + y = 200\), and solved for \(y\) to get \(y = 200 - x\). This expression was then substituted into the total cost equation \(4.95x + 2.65y = 790\).

After substituting, we obtained a single equation \(4.95x + 2.65(200 - x) = 790\), which allowed us to solve for \(x\) alone. Once \(x\) was found, we substituted it back into \(y = 200 - x\) to find \(y\).

This method is efficient and manageable when dealing with two equations, as it simplifies the process to basic algebra.

Elimination Method

The elimination method, also known as the addition or subtraction method, is another technique often used to solve systems of linear equations. This method involves either adding or subtracting the equations directly so that one variable cancels out. It's particularly useful when the coefficients of one of the variables are easily manipulated to match.

In the exercise we explored, Wayne could have used the elimination method by aligning the equations to cancel out one variable. However, given the specific numbers and context, the substitution method was more straightforward.

To use the elimination method, you typically:

• Multiply one or both equations by a constant to align one variable.
• Add or subtract the equations to eliminate one variable.
• Solve the resulting equation for the remaining variable.
• Substitute back to find the other variable.

The elimination method is advantageous when your equations are already arranged in a manner conducive to quick cancellation.

Word Problems

Word problems can often seem tricky, but they are just real-life scenarios modeled with mathematical expressions. The key lies in effectively translating the words into equations.

For Wayne's coffee blending task, the challenge was to correctly understand the relationships between different quantities and costs. This involved setting up proper equations:

• One for the total weight, since we needed to mix 200 pounds in total.
• Another for the total cost, ensuring the blend matches the desired price of \(\$3.95\) per pound.

Strategies for tackling word problems include:

• Reading the problem thoroughly to understand what is being asked.
• Identifying variables and writing them down to clarify the problem.
• Formulating equations based on the relationships and conditions given.
• Using known methods, such as substitution or elimination, to solve the equations.

By breaking down the problem into smaller, manageable steps, word problems become less daunting and more approachable.