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

The length of the rectangle is $11$ inches and the width is $6$ inches.
We know that the formula for the perimeter of a rectangle is: $$P = 2l + 2w$$ We know from the problem that the perimeter is $34$ inches and that the length of the rectangle is $5$ inches more than its width. We put all this information together to get the following equation: $$34 = 2(w + 5) + 2w$$ We distribute what is in parentheses: $$34 = 2w + 10 + 2w$$ Combine like terms: $$34 = 4w + 10$$ Subtract $10$ from both sides to get $w$ on one side of the equation: $$24 = 4w$$ Divide both sides by $4$ to isolate $w$: $$w = 6$$ If we know that the length $l$ is $5$ more inches than the width $w$, then we can find $l$ with the following equation: $$l = 6 + 5$$ Add the right-hand side together to get the value for $l$: $$l = 11$$