Problem 190
Question
Hector spent \(\$ 33.20\) at the post office. He bought four times as many \(\$ 0.41\) stamps as \(\$ 0.02\) stamps. How many of each did he buy?
Step-by-Step Solution
Verified Answer
Hector bought 20 \(0.02\) stamps and 80 \(0.41\) stamps.
1Step 1: Define variables
Let the number of \(\$0.02\) stamps be \(x\). Since Hector bought four times as many \(\$0.41\) stamps, let the number of \(\$0.41\) stamps be \(4x\).
2Step 2: Set up the total cost equation
The total cost can be represented by the equation \[0.02x + 0.41(4x) = 33.20\].
3Step 3: Simplify the equation
Simplify the equation to combine like terms: \[0.02x + 1.64x = 33.20\].
4Step 4: Combine like terms
Add the terms with \(x\): \[1.66x = 33.20\].
5Step 5: Solve for x
Divide both sides of the equation by \(1.66\) to find \(x\): \[x = \frac{33.20}{1.66} ≈ 20\].
6Step 6: Find the number of \( \$0.41\) stamps
Since the number of \(0.41\) stamps is \(4x\), calculate \(4x = 4 \times 20 = 80\). Hector bought \(80\) \(0.41\) stamps.
Key Concepts
Simple EquationsVariable DefinitionCombining Like TermsSolving Linear Equations
Simple Equations
In algebra, simple equations are the building blocks for solving problems. A simple equation is a mathematical statement that shows the equality of two expressions. For example, in the problem where Hector spends money on stamps, we use a simple equation to find out how many stamps he bought.
The general form of these equations looks like this: \[a + b = c\].
By defining the variables and translating the word problem into this form, we can easily find a solution. Simple equations are crucial because understanding them is the first step to tackling more complex algebraic problems.
The general form of these equations looks like this: \[a + b = c\].
By defining the variables and translating the word problem into this form, we can easily find a solution. Simple equations are crucial because understanding them is the first step to tackling more complex algebraic problems.
Variable Definition
Defining variables is an essential step in solving algebra word problems. A variable is a symbol, usually a letter like \( x \) or \( y \), that represents an unknown value in an equation.
In our example about Hector and his stamps, we let \( x \) represent the number of \( \$0.02 \) stamps. By clearly defining \( x \), we can then use it to express other quantities in the problem.
Variables help us turn word problems into mathematical equations, making it easier to solve for unknown values. Properly defining variables ensures that our work is organized and understandable.
In our example about Hector and his stamps, we let \( x \) represent the number of \( \$0.02 \) stamps. By clearly defining \( x \), we can then use it to express other quantities in the problem.
Variables help us turn word problems into mathematical equations, making it easier to solve for unknown values. Properly defining variables ensures that our work is organized and understandable.
Combining Like Terms
Combining like terms is a method used to simplify equations. Like terms are terms that have the same variable raised to the same power. By combining them, we can make our equations simpler and easier to solve.
In the given problem, after setting up the equation\(0.02x + 0.41(4x) = 33.20\), we simplify it by combining like terms: \(0.02x + 1.64x = 33.20\).
This helps us reduce the left side of the equation to \(1.66x\). Combining like terms minimizes the number of terms in an equation, making it more straightforward to find solutions.
In the given problem, after setting up the equation\(0.02x + 0.41(4x) = 33.20\), we simplify it by combining like terms: \(0.02x + 1.64x = 33.20\).
This helps us reduce the left side of the equation to \(1.66x\). Combining like terms minimizes the number of terms in an equation, making it more straightforward to find solutions.
Solving Linear Equations
After simplifying the equation, the next step is to solve the linear equation. A linear equation is an equation in which the highest power of the variable is one. In our example, the simplified equation was\(1.66x = 33.20\).
To solve for \( x \), we divide both sides by \( 1.66 \):\(x = \frac{33.20}{1.66} \).
Solving linear equations often involves isolating the variable on one side of the equation. This process may include dividing, multiplying, adding, or subtracting. Understanding how to solve linear equations is fundamental in algebra, as it allows us to find the value of unknown variables efficiently.
To solve for \( x \), we divide both sides by \( 1.66 \):\(x = \frac{33.20}{1.66} \).
Solving linear equations often involves isolating the variable on one side of the equation. This process may include dividing, multiplying, adding, or subtracting. Understanding how to solve linear equations is fundamental in algebra, as it allows us to find the value of unknown variables efficiently.
Other exercises in this chapter
Problem 188
Jason went to the post office and bought both \(\$ 0.41\) stamps and \(\$ 0.26\) postcards and spent \(\$ 10.28\). The number of stamps was four more than twice
View solution Problem 189
Maria spent \(\$ 12.50\) at the post office. She bought three times as many \(\$ 0.41\) stamps as \(\$ 0.02\) stamps. How many of each did she buy?
View solution Problem 191
Hilda has \(\$ 210\) worth of \(\$ 10\) and \(\$ 12\) stock shares. The numbers of \(\$ 10\) shares is five more than twice the number of \(\$ 12\) shares. How
View solution Problem 192
Mario invested \(\$ 475\) in \(\$ 45\) and \(\$ 25\) stock shares. The number of \(\$ 25\) shares was five less than three times the number of \(\$ 45\) shares.
View solution