A rectangle's length is 3 more than twice its width. If the perimeter of the rectangle is 54, what is the width?

Let $w$ be the width. The length is 3 more than twice the width, so $\ell = 2w + 3$. The perimeter is $2(\ell + w) = 54$, so $2(2w + 3 + w) = 54$, which simplifies to $2(3w + 3) = 54$ and $6w + 6 = 54$. Then $6w = 48$ and $w = 8$, choice B. Check: length 19, and $2(8 + 19) = 54$. Choice A (10) fails the check, giving a perimeter of 66. Choice C (19) is the length, not the width. Choice D (6) fails as well, giving a perimeter of 42. Always confirm which dimension the question asks for.