Problem 219 - Elementary Algebra

Question

In the following exercises, translate to a system of equations and solve. Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Step-by-Step Solution

Answer

Width: 10 feet, Length: 25 feet

Step 1: Understand the Problem

First, identify what we need to find: the length and width of Wayne's patio. The total length of the string of lights, which goes around three sides of the patio, is 45 feet.

Step 2: Define Variables

Let the width of the patio be denoted by the variable \( w \), and the length by the variable \( l \).

Step 3: Create the Equations

Wayne’s patio has three sides surrounded by lights (two widths and one length). This gives the equation: \[ 2w + l = 45 \] Also, the length of the patio is five feet longer than twice its width. This gives the equation: \[ l = 2w + 5 \]

Step 4: Substitute and Solve

Substitute the second equation into the first equation to find \( w \): \[ 2w + (2w + 5) = 45 \]Combine like terms: \[ 4w + 5 = 45 \]Subtract 5 from both sides: \[ 4w = 40 \]Divide by 4: \[ w = 10 \]

Step 5: Solve for Length

Now, substitute the value of \( w \) back into the equation for \( l \): \[ l = 2(10) + 5 \]Simplify: \[ l = 20 + 5 \]So, \( l = 25 \).

Step 6: State the Solution

The width of the patio is 10 feet, and the length is 25 feet.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In word problems, we often translate a scenario into a system of equations to find unknown values. For Wayne's patio problem, we derive two equations representing the relationships given in the problem.

The first equation comes from the total length of the string lights around three sides of the patio. We know this length to be 45 feet and the patio is rectangular, so one length side and two width sides are surrounded by lights. Therefore, the equation becomes:

2w + l = 45

The second equation is derived from the given relationship between the patio's length and width. It states that the length is five feet longer than twice the width, hence:

l = 2w + 5

Together, these equations form a system of equations which will help us solve for both the length and the width of the patio.

Substitution Method

The substitution method is a technique to solve a system of equations. Here’s how it works: we solve one equation for one of the variables, and then substitute that expression into the other equation. This reduces the system to a single equation with one variable.

In Wayne’s problem, we use this method because the second equation is already solved for length ( l = 2w + 5 ), making it straightforward to substitute into the first equation.

We first substitute l from the second equation into the first equation:

2w + (2w + 5) = 45

Now it’s simplified into one variable. We then combine like terms:

4w + 5 = 45

Solving for w :

4w = 40
w = 10

With w known, we substitute it back into the second equation to find l :

l = 2(10) + 5
l = 25

This process shows how substitution can efficiently find the values for both variables.

Rectangular Dimensions

Understanding rectangular dimensions is crucial in solving problems related to geometry and algebra. A rectangle is defined by its length and width. In Wayne's problem, the string lights wrap around three sides of the patio, covering one length and two widths.

To form a clear picture:

• The length ( l ) is the longer side along the house.
• The width ( w ) is the shorter side extending from the house.

The relationship given in Wayne's patio dimensions is between the length and width. Knowing the length is five feet longer than twice the width guides us in creating our equations. We use the perimeter concept here:

• Perimeter of three sides = 2 widths + 1 length
• From the problem: 2w + l = 45

With a solid grasp of the rectangular dimensions, solving any problem relating to geometry becomes simpler.