Problem 47 - Algebra for College Students

Question

Solve each of Problems \(47-62\) by setting up and solving an appropriate algebraic equation. Suppose that the length of a certain rectangle is 2 meters less than four times its width. The perimeter of the rectangle is 56 meters. Find the length and width of the rectangle.

Step-by-Step Solution

Answer

Width is 6 meters and length is 22 meters.

Step 1: Understand the Problem

We need to find the length and width of a rectangle. We're given that the length is 2 meters less than four times the width, and the perimeter of the rectangle is 56 meters.

Step 2: Define Variables

Let's represent the width of the rectangle as \(w\) (in meters). Then the length can be expressed in terms of \(w\) using the given relation: length \(= 4w - 2\).

Step 3: Set Up the Perimeter Equation

The formula for the perimeter of a rectangle is \(P = 2 imes ( ext{length} + ext{width})\). Substituting the given information, we have: \(2 imes ((4w - 2) + w) = 56\).

Step 4: Simplify the Equation

First, simplify the expressions inside the parentheses: \((4w - 2) + w = 5w - 2\). Substitute back into the perimeter equation: \(2 imes (5w - 2) = 56\).

Step 5: Solve for Width

Distribute the 2 to simplify: \(10w - 4 = 56\). Solve for \(w\) by adding 4 to both sides: \(10w = 60\). Then divide by 10: \(w = 6\).

Step 6: Find the Length Using Width

Substitute \(w = 6\) into the expression for length: \(4w - 2 = 4(6) - 2 = 24 - 2 = 22\). Thus, the length is 22 meters.

Key Concepts

Algebraic Equations

To solve problems that involve finding unknown dimensions, like that of a rectangle, we often use algebraic equations. An algebraic equation is an expression that contains variables and constants structured in such a way that they represent a mathematical relationship. In this case, the problem gives relationships between the length and width of the rectangle using an equation. Given that the length is 2 meters less than four times the width, we can express this relationship as:

• Length = \(4w - 2\), where \(w\) is the width.

This equation is crucial because it translates the word problem into a mathematical expression. By turning the problem into an algebraic equation, you create a pathway to find the unknown dimensions using logical mathematical operations. Remember, understanding how to write and manipulate these equations is key to solving such problems.

Perimeter Calculation

Perimeter calculation involves finding the total distance around a shape. For a rectangle, the perimeter is the sum of all its sides. Mathematically, this is represented as:\[ P = 2 \times (\text{length} + \text{width}) \]In our rectangle problem, we know the perimeter is given as 56 meters. By substituting the expressions for the length and width, you form a new equation:\[ 2 \times ((4w - 2) + w) = 56 \]Calculating perimeter in algebra problems often requires substituting known values and expressions into the formula, then simplifying. It's a straightforward way to use given data to solve for unknowns, like the width or length here. Understanding perimeter formulas allows you to tackle similar geometry problems efficiently.

Variable Substitution

Variable substitution is a useful algebraic technique where you replace variables with numbers or other expressions to solve equations. In our rectangle problem, after defining the variables, you reach a point where you need to solve for \(w\), the width. Following the calculations, you have:\[ 2 \times (5w - 2) = 56 \]Simplifying and solving gives:

• Distribute: \(10w - 4 = 56\)
• Add 4 to both sides: \(10w = 60\)
• Solve for \(w\): \(w = 6\)

Once you know the width, substitute \(w = 6\) back into the expression for the length:\[ 4w - 2 = 4(6) - 2 = 22 \]This substitution step is crucial because it allows you to find other related quantities, like the length here, making it a practical tool in many algebraic contexts.