/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 206 Nancy bought seven pounds of ora... [FREE SOLUTION] | 91Ó°ÊÓ

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Nancy bought seven pounds of oranges and three pounds of bananas for \(\$ 17\). Her husband later bought three pounds of oranges and six pounds of bananas for \(\$ 12 .\) What was the cost per pound of the oranges and the bananas?

Short Answer

Expert verified
The cost per pound is \(\textbackslash(O = 2\textbackslash),\textbackslash(B = 1\textbackslash)\)

Step by step solution

01

- Define Variables

Let the cost per pound of oranges be represented by \( O \), and the cost per pound of bananas be represented by \( B \).
02

- Set Up Equations

Using the information given: 1) For Nancy: \[ 7O + 3B = 17 \] 2) For her husband: \[ 3O + 6B = 12 \]
03

- Simplify the Second Equation

Divide the second equation by 3: \[ O + 2B = 4 \]
04

- Solve the Simplified System

Using the equations from Steps 2 and 3, solve for \( O \) and \( B \): First, rewrite the equations: 1) \[ 7O + 3B = 17 \] 2) \[ O + 2B = 4 \] Now, solve equation 2 for \( O \): \[ O = 4 - 2B \] Substitute \( O = 4 - 2B \) into equation 1: \[ 7(4 - 2B) + 3B = 17 \] This simplifies to: \[ 28 - 14B + 3B = 17 \] \[ 28 - 11B = 17 \] \[ -11B = 17 - 28 \] \[ -11B = -11 \] \[ B = 1 \]
05

- Substitute Back to Find \( O \)

Now that \( B = 1 \), substitute back into \( O + 2B = 4 \): \[ O + 2(1) = 4 \] \[ O + 2 = 4 \] \[ O = 2 \]

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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 mathematical expressions that represent straight lines when graphed. They have the general form of \( ax + by = c \), where \(a\), \(b\), and \(c\) are constants. These equations model relationships with two variables, like the cost per pound of oranges and bananas in this problem.

For example, in the problem, Nancy's purchase can be represented by a linear equation:
\(7O + 3B = 17\)
Here:
  • \(O\) represents the cost per pound of oranges
  • \(B\) represents the cost per pound of bananas
Linear equations help break down complex problems into manageable parts by showing relationships between variables.

Understanding linear equations is key in this exercise because they allow us to form the basis of our solution, helping us further work with these values efficiently.
Substitution Method
The substitution method is a technique for solving a system of equations. This method involves solving one of the equations for one variable and then substituting that expression into the other equation.

Using our problem as an example, we start with two linear equations:
  • \(7O + 3B = 17\)
  • \(O + 2B = 4\)
Our goal is to isolate one variable and replace it in the other equation. This is how we do it:
Step 1: Solve \(O + 2B = 4\) for \(O\):
\(O = 4 - 2B\).
Step 2: Substitute \(O = 4 - 2B\) into the first equation:
\(7(4 - 2B) + 3B = 17\)
Step 3: Simplify and solve:
\(28 - 14B + 3B = 17\)
\(-11B + 28 = 17\)
\(-11B = -11\)
\(B = 1\)
Now we substitute \(B\) back to find \(O\):
\(O + 2(1) = 4\)
\(O = 2\).

The substitution method is powerful because it systematically reduces the number of variables, making the equations easier to solve.
Simultaneous Equations
Simultaneous equations are sets of equations with multiple variables that are solved together. Each equation in the system applies constraints to the variables, and the solution is where these constraints overlap.

In our problem, simultaneous equations are used to solve for the prices of oranges and bananas. We started with:
  • \(7O + 3B = 17\)
  • \(3O + 6B = 12\)
These equations reflect the total cost of fruits bought by Nancy and her husband. To solve them, we can either use the substitution method, as seen earlier, or other methods like the elimination method.

Simultaneous equations ensure that the solution satisfies all given conditions, which is essential in many real-world problems. By solving these equations, you can determine the unique values of the variables (\(O\) and \(B\) in our case) that meet all given criteria.
Problem-Solving Steps
Solving algebraic word problems involves a structured approach. Here’s a simple step-by-step process using our exercise as an example:

1. **Understand the Problem**: Read the problem carefully to determine what is being asked. Identify the unknowns—here, the cost per pound of oranges (\(O\)) and bananas (\(B\)).
2. **Define Variables**: Assign symbols to the unknowns.\
3. **Set Up Equations**: Use the information given to form linear equations. For instance:
\(7O + 3B = 17\) and \(3O + 6B = 12\).
4. **Simplify if Necessary**: Simplify equations to make solving easier. Here, divide the second equation by 3 to get:
\(O + 2B = 4\).
5. **Solve the System**: Use methods like substitution to find the solution. Substitute one equation into the other:
\(O = 4 - 2B\) into \(7O + 3B = 17\), leading to:
\(B = 1\), then substitute back to find
\(O = 2\).
6. **Check Your Solution**: Substitute the found values back into the original equations to verify correctness.

Following these problem-solving steps can help you approach algebra problems methodically, ensuring you solve them accurately and efficiently.

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Most popular questions from this chapter

In the following exercises, translate to a system of equations and solve. Marissa wants to blend candy selling for \(\$ 1.80\) per pound with candy costing \(\$ 1.20\) per pound to get a mixture that costs her \(\$ 1.40\) per pound to make. She wants to make 90 pounds of the candy blend. How many pounds of each type of candy should she use?

In the following exercises, translate to a system of equations and solve. Tickets for a dance recital cost \(\$ 15\) for adults and \(\$ 7\) for children. The dance company sold 253 tickets and the total receipts were \(\$ 2,771\). How many adult tickets and how many child tickets were sold?

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