Let's denote the width of the rectangular playing field as $w$ (in yards). According to the problem, the length $l$ of the field is 7 yards longer than triple the width, which can be written as:

$l = 3w + 7$

The perimeter $P$ of a rectangle is given by the formula:

$P = 2l + 2w$

We are told that the perimeter is 422 yards, so:

$2l + 2w = 422$

Now, substitute $l = 3w + 7$ into the perimeter equation:

$2(3w + 7) + 2w = 422$

Simplify this equation:

$6w + 14 + 2w = 422$

Combine like terms:

$8w + 14 = 422$

Subtract 14 from both sides:

$8w = 408$

Divide by 8:

$w = 51$

Now, substitute $w = 51$ into the equation for the length:

$l = 3(51) + 7 = 153 + 7 = 160$

Thus, the dimensions of the rectangular playing field are:

• Width: 51 yards
• Length: 160 yards

Would you like further details or explanations?

Here are 5 related questions to consider:

1. How would the formula for perimeter change if the shape were a square?
2. What if the perimeter were doubled—how would the dimensions change?
3. How would the equation change if the width were 7 yards longer than triple the length?
4. What is the area of this rectangular field?
5. How do you find the diagonal of a rectangular field?

Tip: For any rectangle, the perimeter formula always involves doubling both the length and the width, making it a useful starting point for solving such problems.