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In the following exercises, translate to a system of equations and solve. A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

Short Answer

Expert verified
The length and width are both 15 inches.

Step by step solution

01

- Define Variables

Let the width be represented by \( w \) and the length be represented by \( l \).
02

- Translate Perimeter Information

The perimeter of a rectangle is given by the formula \( P = 2l + 2w \). Given the perimeter is 60 inches, we can write the equation: \[ 2l + 2w = 60 \]
03

- Translate Length Information

According to the problem, the length is fifteen less than twice the width. This can be written as: \[ l = 2w - 15 \]
04

- Substitute and Solve for Width

Substitute the expression for \( l \) from Step 3 into the equation from Step 2: \[ 2(2w - 15) + 2w = 60 \]Simplify and solve for \( w \): \[ 4w - 30 + 2w = 60 \] \[ 6w - 30 = 60 \] \[ 6w = 90 \] \[ w = 15 \]
05

- Solve for Length

Use the value of \( w \) found in Step 4 to solve for \( l \) using the equation from Step 3: \[ l = 2w - 15 \] \[ l = 2(15) - 15 \] \[ l = 30 - 15 \] \[ l = 15 \]
06

- Verify the Solution

Substitute \( l \) and \( w \) back into the perimeter equation to ensure they satisfy the original condition: \[ 2(15) + 2(15) = 60 \] \[ 30 + 30 = 60 \] This confirms that the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

defining variables
One of the first steps in solving any word problem is defining the variables. In this case, we need to determine what quantities we are dealing with. For the problem related to the rectangular family portrait frame, let's define the width of the frame as \( w \) and the length of the frame as \( l \). These variables will help us translate the given information into mathematical equations.

Choosing clear variables is essential:
  • It makes your equations easier to work with.
  • It helps you keep track of what each variable represents.
By defining the width as \( w \) and the length as \( l \), we can proceed to translate the word problem into a system of equations.
translating word problems
Translating word problems into mathematical equations involves understanding both the given information and the relationships between different quantities. For the rectangular frame problem, the essential pieces of information are:
  • The perimeter of the rectangle is 60 inches.
  • The length is fifteen inches less than twice the width.
First, we use the perimeter formula for a rectangle: \( P = 2l + 2w \). Given that the perimeter is 60 inches, we can write:

\[ 2l + 2w = 60 \]

To translate the information about the length, we note that the length is fifteen inches less than twice the width, which can be written as:

\[ l = 2w - 15 \]Now, we have two equations that form our system of equations:
  • \( 2l + 2w = 60 \)
  • \( l = 2w - 15 \)
These equations can now be solved simultaneously to find the values of \( l \) and \( w \).
solving linear equations
Solving systems of linear equations typically involves one of the following methods: substitution, elimination, or graphing. In this exercise, we'll use the substitution method. First, we solve one equation for one variable; in this case, we already have:

\[ l = 2w - 15 \]

Next, we substitute this expression for \( l \) into the perimeter equation:

\[ 2(2w - 15) + 2w = 60 \]

Distribute and combine like terms:

\[ 4w - 30 + 2w = 60 \]

Simplify:

\[ 6w - 30 = 60 \]

\[ 6w = 90 \]

Divide by 6:

\[ w = 15 \]

Now that we have \( w \), use it to find \( l \):

\[ l = 2(15) - 15 \]

\[ l = 30 - 15 \]

\[ l = 15 \]. We've found the width (15 inches) and length (15 inches) of the frame.
perimeter of rectangles
Understanding the perimeter formula is crucial when dealing with rectangular shapes. The perimeter (\( P \)) of a rectangle is the total distance around the edges. The formula includes the lengths of all four sides: two lengths (\( l \)) and two widths (\( w \)).

The perimeter formula is:

\[ P = 2l + 2w \]

In our exercise, the perimeter of 60 inches was used to form the equation:

\[ 2l + 2w = 60 \]. By knowing how to use this formula, we could set up our initial equation and proceed to find the solutions for \( l \) and \( w \).

It’s essential to remember this straightforward perimeter formula, as it frequently appears in various geometry problems. Whether solving for perimeter, width, or length, this formula provides a solid foundation for solving such problems.

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