Problem 59 - Algebra 2

Question

The perimeter of a rectangular picture is 86 inches. Twice the width exceeds the length by 2 inches. What are the dimensions of the picture?

Step-by-Step Solution

Answer

The width is 15 inches and the length is 28 inches.

Step 1: Understand the Problem

We need to find the dimensions (length and width) of a rectangular picture given its perimeter and a relationship between its length and width.

Step 2: Define Variables

Let's denote the width by \( w \) and the length by \( l \). According to the problem: 1. The perimeter is 86 inches.2. Twice the width exceeds the length by 2 inches, i.e., \( 2w = l + 2 \).

Step 3: Write the Perimeter Equation

The perimeter \( P \) of a rectangle can be calculated with the formula: \( P = 2l + 2w \). Substitute the given perimeter: \[ 2l + 2w = 86 \]

Step 4: Simplify Perimeter Equation

Divide the perimeter equation by 2 to simplify it:\[ l + w = 43 \] This equation relates the length and width.

Step 5: Use Length-Width Relationship

Substitute \( l = 2w - 2 \) from the relationship into the simplified perimeter equation:\[ 2w - 2 + w = 43 \]

Step 6: Solve for Width

Combine like terms:\[ 3w - 2 = 43 \]Add 2 to both sides:\[ 3w = 45 \]Divide by 3:\[ w = 15 \]

Step 7: Solve for Length

Using \( l = 2w - 2 \), substitute \( w = 15 \) into the equation:\[ l = 2(15) - 2 = 30 - 2 = 28 \]

Step 8: Verify the Solution

Verify by checking both the perimeter and length-width conditions:- Perimeter: \( 2(28) + 2(15) = 56 + 30 = 86 \)- Length-Width condition: \( 2(15) = 28 + 2 \) which simplifies to \( 30 = 30 \). Both conditions satisfied, thus the solution is correct.

Key Concepts

Understanding the Perimeter Equation

To solve for the dimensions of a rectangular picture, we need to grasp the concept of a perimeter equation. Perimeter, in a simple sense, is the total distance around a two-dimensional shape. For a rectangle, which has opposite sides that are equal, the perimeter \( P \) can be calculated using the formula:\[P = 2l + 2w\]Here, \( l \) represents the length, and \( w \) represents the width of the rectangle. In our problem, the perimeter of the rectangle is given as 86 inches. Therefore, substituting this into the formula, we have:\[2l + 2w = 86\]This equation tells us how the total length around the rectangle relates to its length and width. It's the starting point for further calculations.

Applying Variable Substitution

Variable substitution is a technique we use to simplify equations when we have multiple unknowns. In this exercise, we know that twice the width exceeds the length by 2 inches. This relationship can be mathematically expressed as:\[2w = l + 2\]This equation allows us to express the length \( l \) in terms of width \( w \). Rewriting, we get:\[l = 2w - 2\]Now, we have two equations:

• The simplified perimeter equation: \( l + w = 43 \)
• And the length-width relation: \( l = 2w - 2 \)

By substituting \( l = 2w - 2 \) into the perimeter equation, we eliminate one variable. This substitution technique is crucial because it lets us focus on finding the width first, which we can then plug back to find the length.

Solving Linear Equations

With the equations set up and the substitution done, our task now is to solve the linear equation for one of the unknowns. Substituting into our simplified perimeter equation, we have:\[2w - 2 + w = 43\]Combine like terms:\[3w - 2 = 43\]Next, to isolate \( w \), add 2 to both sides:\[3w = 45\]Divide both sides by 3:\[w = 15\]Now that we've solved for \( w \), which is the width of the rectangle, we use the relationship \( l = 2w - 2 \) to find the length:\[l = 2(15) - 2 = 30 - 2 = 28\]Solving linear equations involves systematically isolating variables to find their values. Here, the equations worked together to give us both the width and the length. Finally, always verify the solutions back through the original problem conditions to ensure they're correct.