Problem 220 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve. Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his rectangular front yard. The length, the side along the house, is five feet less than three times the width. Find the length and width of the fencing.

Step-by-Step Solution

Answer

Width = 41 feet, Length = 118 feet.

Step 1: Define Variables

Let the width of the fencing be denoted as \( w \) and the length of the fencing be denoted as \( l \).

Step 2: Setup the Perimeter Equation

Since Darrin is hanging garland around three sides of the rectangular front yard, which includes two widths and one length, the equation for the perimeter is given by: \( 2w + l = 200 \).

Step 3: Setup the Length-WIDTH Relationship

According to the problem, the length \( l \) is five feet less than three times the width \( w \). Therefore, the second equation can be written as: \( l = 3w - 5 \).

Step 4: Substitute the Second Equation into the First

Replace \( l \) in the first equation with \( 3w - 5 \): \( 2w + (3w - 5) = 200 \).

Step 5: Solve for \( w \)

Combine like terms and solve for \( w \): \[ 2w + 3w - 5 = 200 \] \[ 5w - 5 = 200 \] \[ 5w = 205 \] \[ w = 41 \]

Step 6: Solve for \( l \)

Use the value of the width \( w \) to find the length \( l \): \[ l = 3(41) - 5 = 123 - 5 = 118 \]

Step 7: Conclusion

The width of the fencing is 41 feet and the length of the fencing is 118 feet.

Key Concepts

headline of the respective core concept

Problem solving often requires a systematic approach to break down the issue into manageable parts. In the given exercise, the problem revolves around calculating the dimensions of a rectangular yard using logical steps. Consider beginning by comprehending the problem statement fully.
Determine the knowns and unknowns. For instance, we know the total garland length—200 feet. We also know the relationship between the length and the width of the yard.
Next, transition to defining variables. We use algebra to represent the unknown dimensions. We let the width be denoted by \( w \) and the length by \( l \).
Setting up a system of equations follows. The problem mentions the use of garland around three sides with the equation: \( 2w + l = 200 \). Meanwhile, the length is defined as: \( l = 3w - 5 \).
Lastly, solve the system step-by-step. This involves substituting one equation into another, solving for one variable, then using that to find the other. Ensure each step logically flows to minimize errors and enhances understanding.

headline of the respective core concept

Algebraic expressions allow us to describe mathematical relationships. In this problem, the expressions relate to the yard's dimensions using the width (\( w \)) and the length (\( l \)).
We write the perimeter equation as: \( 2w + l = 200 \). This expression encapsulates the total garland used, with \( 2w \) for the width sides and \( l \) for the length side.
An additional algebraic expression describes how the yard’s length relates to its width: \( l = 3w - 5 \). Here, 3 times the width minus 5 gives the length of the yard.
The power of algebraic expressions lies in their ability to convert word problems into solvable mathematical equations. By expressing the problem variables algebraically, solving becomes a straightforward process through substitution and basic algebraic manipulations.

headline of the respective core concept

The substitution method is a key technique for solving systems of equations. Here's how it’s applied in this problem:
We start with the two equations: \( 2w + l = 200 \) and \( l = 3w - 5 \). The substitution method involves solving one equation for one variable and substituting this into the other equation.
First, we solve the second equation for \( l \): \( l = 3w - 5 \).
Next, substitute \( l \) into the first equation: \( 2w + (3w - 5) = 200 \). This replaces \( l \) with its equivalent expression in terms of \( w \).
Simplify and solve for \( w \):
\[ 2w + 3w - 5 = 200 \]
\[ 5w - 5 = 200 \]
\[ 5w = 205 \]
\[ w = 41 \]
With \( w \) found, substitute back to find \( l \): \( l = 3(41) - 5 = 118 \).
The substitution method streamlines solving the system by focusing on one variable at a time and is especially useful in problems with clear variable relationships.