Problem 38 - Intermediate Algebra

Question

U.S. Currency. The perimeter of a one-dollar bill is 17.5 inches and the length is 0.92 in. more than twice the width. Find the dimensions of a one-dollar bill.

Step-by-Step Solution

Answer

The width is 2.61 inches and the length is 6.14 inches.

Step 1 - Define Variables

Let the width of the one-dollar bill be represented as \( w \). According to the problem, the length of the bill, \( l \), is given as \( l = 2w + 0.92 \).

Step 2 - Express Perimeter Equation

The perimeter of a rectangle is calculated by the formula \( P = 2l + 2w \). We know the perimeter is 17.5 inches, so we can set up the equation \( 2l + 2w = 17.5 \).

Step 3 - Substitute Length Equation

Substitute the equation for length from Step 1 into the perimeter equation: \( 2(2w + 0.92) + 2w = 17.5 \).

Step 4 - Simplify and Solve for Width

Simplify the equation: \( 4w + 1.84 + 2w = 17.5 \). Combine like terms: \( 6w + 1.84 = 17.5 \). Subtract 1.84 from both sides to get \( 6w = 15.66 \). Divide by 6: \( w = 2.61 \).

Step 5 - Solve for Length

Use the value for width to find the length: \( l = 2(2.61) + 0.92 = 5.22 + 0.92 = 6.14 \).

Key Concepts

Problem Solving in Algebra

Tackling algebraic equations can initially feel intimidating, but embracing a methodical approach can simplify the process significantly. Let’s look at a typical problem-solving strategy when dealing with algebraic equations, especially when finding unknown dimensions like those of a one-dollar bill.
Start by clearly understanding the problem statement. Identify what is being asked. Here, we need to find the dimensions of the bill using given clues.
Next, form a plan: define what needs to be found and what you know. In our exercise, we recognize:

• The perimeter is known (17.5 inches), and
• The relationship between length and width is given.

By defining variables (in this case, the width and length), you create a foundation to build equations that represent real-world relationships.
This skill not only aids in algebra but extends to everyday problem-solving, helping to methodically break down complex problems.

Understanding Variables

Variables are symbols that stand in for unknown or changeable numbers. In mathematics, they are essential for writing equations and expressions that reflect the conditions of a problem.
In our exercise, we let the variable \( w \) represent the width of the bill, because this is our unknown quantity. The length, expressed as \( l = 2w + 0.92 \), also involves a variable since it depends on the width.
This conceptualization of \( l \) and \( w \) helps translate a word problem into mathematical language.
Using variables allows us to write and solve equations to find these unknowns effectively, showing the powerful role they play in algebraic problem-solving. They act as placeholders that we manipulate using the given information to ultimately solve the equation for specific values.

Perimeter Calculation

The perimeter of a rectangle is the total distance around the outer edges. Calculating it necessitates understanding the formula: \[ P = 2l + 2w \].
This formula sums up twice the length and twice the width, since a rectangle has opposite sides that are equal in length.
In this exercise, we were given the total perimeter as 17.5 inches. Therefore, we set up an equation:
\[ 2l + 2w = 17.5 \].
Calculating the perimeter is often a key step in solving problems that require you to find missing measurements, such as dimensions of shapes.
By using measurements you can determine from known values, perimeter equations serve as a practical tool in both academic and real-world settings.

Rectangular Dimensions

Rectangular dimensions refer to the length and width of a rectangle. Since a rectangle has opposite sides that are equal, it’s crucial to understand how these dimensions relate to its perimeter and area.
In this context, the one-dollar bill's dimensions are deduced from an algebraic equation where the length (\( l \)) is noted as slightly more than twice the width (\( w \)). This relationship can be expressed as: \[ l = 2w + 0.92 \].
Using known equations and perimeter, we solve for \( w \) and substitute back to find \( l \). Calculated dimensions from our exercise are approximately 2.61 inches for width and 6.14 inches for length.
Understanding these dimensions aids in visualizing and working with physical attributes of rectangular shapes, a skill beneficial not only in mathematics but also in design and engineering tasks.