Problem 87 - Elementary Algebra

Question

Set up an algebraic equation and use it to solve the following. The length of a rectangle is 6 times its width. If the area is 96 square inches, then find the dimensions of the rectangle.

Step-by-Step Solution

Answer

The dimensions of the rectangle are 4 inches (width) and 24 inches (length).

Step 1: Identify Variables

Let's define the variables for the problem. Let \( w \) be the width of the rectangle. Since the length is 6 times the width, let \( l = 6w \) be the length.

Step 2: Formulate the Equation

The area of a rectangle is calculated as \( ext{Area} = ext{length} imes ext{width} \). According to the problem, the area is 96 square inches. So, we can write the equation as \( l \times w = 96 \). Substitute \( l = 6w \) into the equation to get \( 6w \times w = 96 \).

Step 3: Simplify the Equation

Simplify the equation \( 6w \times w = 96 \) to find the width in terms of a single variable: \( 6w^2 = 96 \).

Step 4: Solve for the Width

To find \( w \), divide both sides of the equation by 6: \( w^2 = 16 \). Then, take the square root of both sides to find \( w = 4 \).

Step 5: Calculate the Length

Since \( l = 6w \), substitute \( w = 4 \) into this equation to find \( l = 6 \times 4 = 24 \).

Step 6: Verify the Solution

Double-check that the dimensions meet the area requirement: \( l \times w = 24 \times 4 = 96 \). Thus, the solution is correct.

Key Concepts

Rectangular Area

Understanding the concept of rectangular area is crucial when tackling problems involving dimensions of a rectangle. A rectangle is a four-sided polygon where opposite sides are equal in length. The area of a rectangle is the amount of space it occupies, which can be calculated by multiplying its length by its width. In formula terms, this is represented as:

• Area = Length × Width

To find the area, simply measure the rectangle's length and width, then multiply these two values. In many real-world problems, you might be given the area and one dimension, needing to work backward to find the missing dimension. This requires setting up and solving algebraic equations based on the area formula.

Variables in Algebra

Variables in algebra are symbols or letters, like \( w \) or \( l \), used to represent unknown values. They are fundamental tools that help in forming equations and solving problems. In our rectangle problem, variables allow us to define and manipulate the dimensions effectively:

• \( w \) stands for the width of the rectangle.
• \( l = 6w \) represents the length as it is 6 times the width.

By using variables, complex relationships can be simplified into one or more equations. This method helps eliminate the unknowns step-by-step through substitution or other algebraic techniques. Variables simplify the process of deriving solutions to problems by expressing relationships clearly and concisely.

Solving Quadratic Equations

Quadratic equations come into play when dealing with expressions that include squares of unknowns, like \( w^2 \). The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Solving these equations requires specific techniques, often involving:

• Factorization, where you express the quadratic as a product of two binomials.
• The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which provides the solutions directly.
• Completing the square, a method used to transform the quadratic into a perfect square trinomial.

For the rectangle problem, we derived a simpler quadratic equation \( 6w^2 = 96 \). Solving this involved first isolating \( w^2 \) by dividing by 6, then taking the square root to find \( w = 4 \). Recognizing and solving quadratics is vital as they frequently appear in geometric, algebraic, and real-world applications.