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Chapter 6: Problem 8

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. A storekeeper is preparing a mixture of peanuts and raisins. If peanuts cost \(\$ 1.60\) per pound and raisins cost \(\$ 1.85\) per pound, how many pounds of each should be used to prepare 50 pounds of a mixture selling at \(\$ 1.70\) per pound?

Short Answer

Expert verified
Use 30 pounds of peanuts and 20 pounds of raisins.

Step by step solution

01

- Define the variables

Let \(x\) represent the number of pounds of peanuts and \(y\) represent the number of pounds of raisins. We need to find the values of \(x\) and \(y\).
02

- Set up the equations based on the given conditions

We have two main conditions to form the equations: 1. The total weight of the mixture is 50 pounds: \(x + y = 50\).2. The total cost of the mixture at \( \$1.70 \) per pound: \(1.60x + 1.85y = 1.70 \times 50\).
03

- Simplify the cost equation

First, simplify the second equation:\(1.60x + 1.85y = 85\).
04

- Solve one equation for one variable

From the first equation, solve for \(y\) in terms of \(x\):\(y = 50 - x\).
05

- Substitute into the second equation

Substitute \(y = 50 - x\) into the second equation:\(1.60x + 1.85(50 - x) = 85\).
06

- Solve for \(x\)

Distribute and combine like terms to solve for \(x\): \(1.60x + 92.5 - 1.85x = 85\) \(-0.25x + 92.5 = 85\) \(-0.25x = -7.5\) \(x = 30\).
07

- Solve for \(y\)

Use \(x = 30\) to find \(y\): \(y = 50 - x\) \(y = 50 - 30\) \(y = 20\).
08

- Verify the solution

Substitute \(x = 30\) and \(y = 20\) back into the original equations to ensure they are correct:\(x + y = 50\) and \(1.60x + 1.85y = 85\). \(30 + 20 = 50\) \(1.60(30) + 1.85(20) = 85\). Both equations are satisfied, so the solution is verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Equations
Linear equations are equations where each term is either a constant or the product of a constant and a single variable. In this problem, we use two linear equations to represent the total weight and total cost of a peanut and raisin mixture. The first equation, derived from the given total weight, is written as:
  • \(x + y = 50\)
Here, \(x\) and \(y\) represent the weights of peanuts and raisins, respectively. The second equation is based on the cost of the mixture:
  • \(1.60x + 1.85y = 85\)
Linear equations are crucial in mixture problems because they help express relationships between quantities in a simple form.
Mixture Problems
Mixture problems involve combining different substances to form a mixture with a specific goal. In this case, the goal is to mix peanuts and raisins to create a 50-pound mixture that sells for \(\text{\$1.70}\) per pound. Key steps include identifying the substances and their respective costs, determining the total weight and price of the mixture, and setting up equations that reflect these conditions. The key equations in our mixture problem are:
  • \(x + y = 50\) (total weight)
  • \(1.60x + 1.85y = 85\) (total cost)
Substitution Method
The substitution method is a technique for solving systems of linear equations. First, we solve one of the equations for a single variable. In our problem, we solve the first equation for \(y\):
  • \(y = 50 - x\)
Next, we substitute this expression for y into the second equation:
  • \(1.60x + 1.85(50 - x) = 85\)
By substituting, we transform a system of two equations with two variables into a single equation with one variable, making it easier to solve.
Cost Analysis
Cost analysis in algebraic problems involves setting up equations based on price per unit. For this exercise, we use the costs of peanuts (\text{\$1.60} per pound) and raisins (\text{\$1.85} per pound) to form our second equation. The combined cost of the peanuts and raisins times their respective weights must equal the desired mixture cost:
  • \(1.60x + 1.85y = 1.70 \times 50 = 85\)
This equation ensures the mixture's total cost exactly matches the cost we want per pound.
Verification of Solutions
Verifying a solution means checking whether it satisfies all given conditions. Upon solving, we found that \(x = 30\) and \(y = 20\). To verify, we substitute these back into our original equations:
  • Total weight: \(30 + 20 = 50\)
  • Total cost: \(1.60 \times 30 + 1.85 \times 20 = 48 + 37 = 85\)
Both conditions are met, confirming that our solution is correct and validating the process we used to solve the problem.

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Most popular questions from this chapter

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{l} 28 x-4.6 y=3.5 \\ 5.2 x-3.4 y=10.1 \end{array}\right.$$

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{l} \frac{x}{3}-\frac{y}{6}=5 \\ \frac{x}{4}-\frac{y}{2}=15 \end{array}\right.$$

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. The perimeter of a rectangle is \(96 \mathrm{ft}\). Three times the width is 8 less than the length. Find the dimensions of the rectangle.

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{aligned} 3 x+6 y &=2 \\ -3 x-3 y &=1 \end{aligned}\right.$$

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{aligned} 11 a-2 b &=30 \\ 3 a+3 b &=-6 \end{aligned}\right.$$
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