Be sure to use proper Word Problem format for the following problems.
4. International basketball competitions are played on a rectangular court where the
length is 2m less than twice the width. If the perimeter of the court is 86m, what are the
dimensions of the court?



Answer :

Certainly! Let's solve the word problem step-by-step.

### Problem
International basketball competitions are played on a rectangular court where:
- The length of the court is 2 meters less than twice the width.
- The perimeter of the court is 86 meters.

We are to find the dimensions (length and width) of the court.

### Step-by-Step Solution

1. Define Variables:
Let [tex]\( w \)[/tex] be the width of the court in meters.
Let [tex]\( l \)[/tex] be the length of the court in meters.

2. Translate the Problem Conditions into Equations:

- According to the problem, the length [tex]\( l \)[/tex] is 2 meters less than twice the width:
[tex]\[ l = 2w - 2 \][/tex]

- The perimeter of a rectangle is given by the formula:
[tex]\[ \text{Perimeter} = 2 \times (\text{length} + \text{width}) \][/tex]

- The perimeter of the court is given as 86 meters:
[tex]\[ 2(l + w) = 86 \][/tex]

3. Substitute the Expression for the Length into the Perimeter Equation:
- Using the equation [tex]\( l = 2w - 2 \)[/tex], substitute [tex]\( l \)[/tex] into the perimeter equation:
[tex]\[ 2((2w - 2) + w) = 86 \][/tex]

4. Simplify and Solve for [tex]\( w \)[/tex]:
[tex]\[ 2(2w - 2 + w) = 86 \][/tex]
[tex]\[ 2(3w - 2) = 86 \][/tex]
[tex]\[ 6w - 4 = 43 \][/tex]
[tex]\[ 6w = 45 \][/tex]
[tex]\[ w = 7.5 \][/tex]

So, the width [tex]\( w \)[/tex] of the court is 7.5 meters.

5. Find the Length [tex]\( l \)[/tex]:
- Substitute [tex]\( w = 7.5 \)[/tex] back into the expression for the length:
[tex]\[ l = 2w - 2 \][/tex]
[tex]\[ l = 2(7.5) - 2 \][/tex]
[tex]\[ l = 15 - 2 \][/tex]
[tex]\[ l = 13 \][/tex]

So, the length [tex]\( l \)[/tex] of the court is 13 meters.

### Conclusion:
The dimensions of the basketball court are:
- Width: 7.5 meters
- Length: 13 meters