High School

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The length of a rectangle is 2 feet less than three times the width. If the perimeter is 68 feet, what are the dimensions of the rectangle?

Equations:

1. [tex] l = 3w - 2 [/tex]

2. [tex] 2l + 2w = 68 [/tex]

Answer :

To solve the problem of finding the dimensions of the rectangle, we need to use the information given about its length and perimeter. Let's break down the problem step-by-step:

1. Define Variables:
- Let [tex]\( l \)[/tex] represent the length of the rectangle.
- Let [tex]\( w \)[/tex] represent the width of the rectangle.

2. Set Up Equations Based on the Problem:
- From the problem, we know that the length [tex]\( l \)[/tex] is 2 feet less than three times the width. This relationship can be expressed as:
[tex]\[
l = 3w - 2
\][/tex]
- We also know that the perimeter of the rectangle is 68 feet. The formula for the perimeter of a rectangle is:
[tex]\[
P = 2l + 2w
\][/tex]
Since the perimeter is given as 68, we can substitute it into the equation:
[tex]\[
2l + 2w = 68
\][/tex]

3. Substitute the Expression for [tex]\( l \)[/tex]:
- From the first equation, we have [tex]\( l = 3w - 2 \)[/tex]. Substitute this expression for [tex]\( l \)[/tex] in the perimeter equation:
[tex]\[
2(3w - 2) + 2w = 68
\][/tex]

4. Simplify and Solve for [tex]\( w \)[/tex]:
- Distribute the 2 in the expression:
[tex]\[
6w - 4 + 2w = 68
\][/tex]
- Combine like terms:
[tex]\[
8w - 4 = 68
\][/tex]
- Add 4 to both sides to isolate the terms with [tex]\( w \)[/tex]:
[tex]\[
8w = 72
\][/tex]
- Divide by 8 to solve for [tex]\( w \)[/tex]:
[tex]\[
w = 9
\][/tex]

5. Find [tex]\( l \)[/tex] Using the Width:
- Now that we have [tex]\( w = 9 \)[/tex], substitute it back into the equation for [tex]\( l \)[/tex]:
[tex]\[
l = 3(9) - 2
\][/tex]
- Calculate the value:
[tex]\[
l = 27 - 2 = 25
\][/tex]

6. Conclusion:
- The dimensions of the rectangle are:
- Length ([tex]\( l \)[/tex]): 25 feet
- Width ([tex]\( w \)[/tex]): 9 feet

These are the dimensions of the rectangle that satisfy the conditions given in the problem.

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