Solve the problem.
-The perimeter of a rectangle is $54 \mathrm{~m}$. If the width were doubled and the length were increased by $17 \mathrm{~m}$, the perimeter would be $104 \mathrm{~m}$. What are the length and width of the rectangle?
A) width $8 \mathrm{~m}$, length $13 \mathrm{~m}$
B) width $8 \mathrm{~m}$, length $19 \mathrm{~m}$
C) width $19 \mathrm{~m}$, length $8 \mathrm{~m}$
D) width $13 \mathrm{~m}$, length $13 \mathrm{~m}$

Question

Quizplus · Accepted Answer

The original perimeter equation is $2l + 2w = 54$. The modified perimeter equation is $2(l + 17) + 2(2w) = 104$. Solving these equations gives width $8 \mathrm{~m}$ and length $19 \mathrm{~m}$.

Solve the Problem 54 m54 \mathrm{~m}54 m If the Width Were Doubled and the Length Were

Solve the Problem $54 \mathrm{~m}$ If the Width Were Doubled and the Length Were