Problem 24 - Intermediate Algebra

Question

The owner of a candy store wants to mix some peanuts worth \(\$ 3\) per pound, some cashews worth \(\$ 9\) per pound, and some Brazil nuts worth \(\$ 9\) per pound to get 50 pounds of a mixture that will sell for \(\$ 6\) per pound. She uses 15 fewer pounds of cashews than peanuts. How many pounds of each did she use?

Step-by-Step Solution

Answer

25 pounds of peanuts, 10 pounds of cashews, and 15 pounds of Brazil nuts.

Step 1: Define Variables

Let \( x \) be the pounds of peanuts, \( y \) be the pounds of cashews, and \( z \) be the pounds of Brazil nuts. We can set up equations based on the problem statement.

Step 2: Set Up Total Weight Equation

The total weight of the mixture is 50 pounds. So, we have:\[ x + y + z = 50 \]

Step 3: Set Up Price Equation

The total cost of the mixture is \\(6 per pound for 50 pounds, which is \\)300. So, we can write:\[ 3x + 9y + 9z = 300 \]

Step 4: Use Cashews and Peanuts Relationship

The owner uses 15 fewer pounds of cashews than peanuts. Therefore:\[ y = x - 15 \]

Step 5: Substitute y in Previous Equations

Substitute \( y = x - 15 \) in the total weight equation:\[ x + (x - 15) + z = 50 \] Simplifying, we get:\[ 2x + z = 65 \] Substitute \( y = x - 15 \) in the price equation:\[ 3x + 9(x - 15) + 9z = 300 \] Simplifying, we get:\[ 12x + 9z = 435 \]

Step 6: Solve for z from Total Weight Equation

From the equation:\[ 2x + z = 65 \], solve for \( z \):\[ z = 65 - 2x \]

Step 7: Substitute for z in Price Equation

Substitute \( z = 65 - 2x \) in the price equation:\[ 12x + 9(65 - 2x) = 435 \] Simplifying, we get:\[ 12x + 585 - 18x = 435 \] Which simplifies further to:\[ -6x + 585 = 435 \]

Step 8: Solve for x

From \[ -6x + 585 = 435 \], subtract 585 from both sides to get:\[ -6x = -150 \] Dividing both sides by -6, we find:\[ x = 25 \]

Step 9: Find y and z

Now that we have \( x = 25 \), find \( y \) using \( y = x - 15 \):\[ y = 25 - 15 = 10 \]Now, find \( z \) using \( z = 65 - 2x \):\[ z = 65 - 2(25) = 15 \]

Step 10: Final Answer Verification

We have \( x = 25 \), \( y = 10 \), and \( z = 15 \). Sum equals 50, and price check:Original price equation was:\[ 3(25) + 9(10) + 9(15) = 75 + 90 + 135 = 300 \]All conditions are satisfied, confirming our solution.

Key Concepts

Mixture Problems

Mixture problems often come up in real-world applications, like our candy store example, where multiple components are combined to create a new product with specific properties. These problems generally involve finding quantities that satisfy certain conditions regarding both weight and cost.

In this exercise, we have three types of nuts: peanuts, cashews, and Brazil nuts. Each type has its own price per pound, and we are tasked with creating a mixture that meets a specific total weight and price per pound.

To solve a mixture problem, first:

• Identify the different components of the mixture.
• Determine any given relationships between the components.
• Set up equations based on these relationships, typically involving total weight and cost.

Mixture problems are often solved using algebraic methods such as creating and solving systems of equations, which we'll discuss next.

Variable Substitution

Variable substitution is a crucial technique in solving equations, especially when dealing with mixture problems or systems of equations. The idea is to replace one variable with an expression involving another variable, simplifying the system into fewer equations.

For instance, in our problem, we know that the owner uses 15 pounds less cashews than peanuts. We express the number of pounds of cashews, \( y \), as \( y = x - 15 \) where \( x \) represents the pounds of peanuts. With this substitution, you can focus on solving equations with fewer unknown variables.

The steps to successfully apply variable substitution include:

• Identify a relationship between variables.
• Express one variable in terms of another.
• Substitute this expression into other equations, simplifying whenever possible.

This technique is vital in reducing the complexity of the problem, as it allows us to solve for remaining unknowns more efficiently.

System of Equations

A system of equations is a collection of two or more equations with a set of variables. Solving a system of equations allows us to find the values of these variables that satisfy all the equations at once.

In our candy store example, we have equations representing both the total weight and cost of the nut mixture, alongside a specific relationship between peanuts and cashews. Here's a summary of how they work together:

• The total weight equation: \( x + y + z = 50 \)
• The cost equation: \( 3x + 9y + 9z = 300 \)
• The relationship between peanuts and cashews: \( y = x - 15 \)

By using substitution and some algebraic manipulations, these equations can be reduced to a simpler form and solved step by step.

The process of solving a system of equations often involves:

• Using substitution to express unknowns with fewer variables.
• Simplifying equations to isolate variables.
• Solving equations step by step until all variable values are found.

Understanding how to work through a system of equations is a powerful skill that applies to many areas beyond just mixture problems.