## Introductory Algebra for College Students (7th Edition)

The length of the field is $200$ yards and the width is $50$ yards.
We know that the perimeter of a rectangle can be given by the following formula: $$P = 2l + 2w$$ We know from the problem that the perimeter is $500$ yards and that the length of the field is four times greater than its width. We can set up an equation as follows: $$500 = 2(4w) + 2w$$ We can simplify this equation: $$8w + 2w = 500$$ Combine like terms: $$10w = 500$$ Solve for $w$ by dividing each side by $10$: $$w = 50$$ We know that the length, $l$, is four times greater than the width; therefore, we can find length with the following equation: $$l = 4(50)$$ Solve for $l$: $$l = 200$$ The length of the field is $200$ yards and the width is $50$ yards.