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Chapter 2: Problem 13

Part of the proceeds from a garage sale was \(\$ 280\) worth of \(\$ 5\) and \(\$ 10\) bills. If there were 20 more \(\$ 5\) bills than \(\$ 10\) bills, find the number of each denomination. $$ \begin{array}{c|c|c} \hline & {\text {Number of Bills }} & {\text {Value of Bills}} \\ \hline \$ 5 \text { bills } & {} \\ \hline \$ 10 \text { bills } \\ \hline \text { Total } \\ \hline \end{array} $$

Short Answer

Expert verified
There are 32 \$5 bills and 12 \$10 bills.

Step by step solution

01

Define Variables

Let \( x \) be the number of \( \\(10 \) bills. According to the problem, there are 20 more \( \\)5 \) bills than \( \\(10 \) bills. Therefore, we can define the number of \( \\)5 \) bills as \( x + 20 \).
02

Formulate Equations

Since the total amount from the \( \\(5 \) and \( \\)10 \) bills is \( \$280 \), we can set up the equation based on their values: \( 5(x + 20) + 10x = 280 \).
03

Simplify the Equation

Substitute the expression for the number of \( \$5 \) bills into the equation: \[ 5(x + 20) + 10x = 280 \]. This simplifies to \[ 5x + 100 + 10x = 280 \].
04

Solve for x

Combine like terms in the equation: \[ 15x + 100 = 280 \]. Subtract 100 from both sides to get \[ 15x = 180 \].
05

Find the Number of $10 Bills

Divide both sides by 15 to solve for \( x \): \[ x = 12 \]. Therefore, there are 12 \( \$10 \) bills.
06

Find the Number of $5 Bills

Since there are 20 more \( \\(5 \) bills than \( \\)10 \) bills, add 20 to the number of \( \\(10 \) bills: \( 12 + 20 = 32 \). So, there are 32 \( \\)5 \) bills.
07

Verify the Solution

Check the total amount calculation: Total from \( \\(5 \) bills is \( 5 \times 32 = 160 \), and total from \( \\)10 \) bills is \( 10 \times 12 = 120 \). The combined total is \( 160 + 120 = 280 \), which matches the given total.

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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 fundamental in solving algebra word problems. In the context of this problem, a linear equation is used to calculate the total value of bills collected from a garage sale. Linear equations often involve expressions where variables represent quantities that can be adjusted to solve the problem. Here, the linear equation is set up as follows:
  • You're adding the total values of two types of bills, the \(5 bills and the \)10 bills, to equal $280.
  • The equation is written as: \[ 5(x + 20) + 10x = 280 \]
Linear equations like this make it possible to solve for unknown variables by combining terms to form a simpler expression, leveling the playing field for all unknowns involved. Understanding how to set up and solve these equations is crucial for handling more complex algebra word problems.
Variable Definition
Defining variables is a critical step in transforming a word problem into a solvable mathematical equation. In this problem, variables help simplify the relationships described in the scenario.
  • We define \( x \) as the number of \(10 bills.
  • Based on the problem, there are 20 more \)5 bills than \(10 bills.
  • This relation is expressed as \( x + 20 \) for the number of \)5 bills.
By clearly defining variables, each aspect of the problem becomes easier to track and solve. This step helps eliminate confusion and guides the formulation of equations. The variable serves as a placeholder for the unknown quantities, which are then solved methodically.
Equation Simplification
Simplifying equations is a vital part of problem-solving in algebra as it transitions complex expressions to simpler, more manageable ones. The goal is to continue simplifying until the equation is easy to solve that it reveals the value of the unknown variables. In our scenario:
  • Begin with the equation: \[ 5(x + 20) + 10x = 280 \]
  • Distribute the 5 into \( (x+20) \) to get \[ 5x + 100 + 10x = 280 \]
  • Combine like terms (the terms involving \( x \)) to transform it into \[ 15x + 100 = 280 \]
  • Subtract 100 from both sides to further simplify: \[ 15x = 180 \]
Each simplification step gradually reduces the complexity of the equation, focusing it towards finding the value of \( x \). These incremental simplifications make the path to the solution straightforward and clear.
Problem Solving in Algebra
Problem-solving in algebra is an essential skill that blends together defining variables, setting up equations, and systematically solving them. Following a sequence of logical steps, students can tackle seemingly complex problems efficiently.
  • Start by understanding the problem鈥攚hat is asked and what is given.
  • Define your variables carefully to ensure they correctly model the problem.
  • Formulate equations based on the relationships described in the problem.
  • Simplify the equations to solve for the unknowns.
  • Always verify your solution by checking if it satisfies the original conditions of the problem.
In this garage sale example, each step methodically leads toward understanding the relationships between the types of bills. Solving such problems enhances computational ability and promotes logical thinking. The systematic process is a crucial part of developing algebraic intuition and efficiency.

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Most popular questions from this chapter

Solve each inequality. Graph the solution set and write it in interval notation. $$ 1<4+2 x \leq 7 $$

A stack of \(\$ 20, \$ 50,\) and \(\$ 100\) bills was retrieved as part of an FBI investigation. There 46 more \(\$ 50\) bills than \(\$ 100\) bills. Also, the number of \(\$ 20\) bills was 7 times the number of \(\$ 100\) bills. If the total value of the money was \(\$ 9550\), find the number of each type of bill.

Fill in the blanks with numbers of your choice so that each equation has the given solution. Note: Each blank may be replaced by a different number. \(x+\quad=2 x-\quad ;\) solution: 9

Explain how solving a linear inequality is similar to solving a compound inequality.

Solve each equation. $$ t^{2}-6 t=t(8+t) $$
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