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

The width of the football field is $53$ meters and the length is $120$ meters.
We know that the perimeter of a rectangle is given by the following formula: $$P = 2l + 2w$$ We know that the perimeter is $346$ meters, We also know that the length is $14$ meters longer than twice the width. If $w$ is the measure of the width, then let $2w + 14$ be the measure of the length of this football field. We can now plug in this information into the formula for perimeter: $$346 = 2(2w + 14) + 2w$$ Use distributive property: $$346 = 4w + 28 + 2w$$ Combine like terms: $$6w + 28 = 346$$ Subtract $28$ from each side of the equation to isolate the variable: $$6w = 318$$ Divide each side by $6$ to solve for $w$: $$w = 53$$ We know that the length $l$ is $14$ more than twice the width $w$, we can use this information to solve for $l$: $$l = 2(53) + 14$$ Multiply first according to order of operations: $$l = 106 + 14$$ Now we can add: $$l = 120$$ The width of the football field is $53$ meters and the length is $120$ meters.