Solve the problem.
-The perimeter of a rectangle is $52 \mathrm{~m}$. If the width were doubled and the length were increased by $24 \mathrm{~m}$, the perimeter would be $114 \mathrm{~m}$. What are the length and width of the rectangle?
A) width $13 \mathrm{~m}$, length $13 \mathrm{~m}$
B) width $19 \mathrm{~m}$, length $7 \mathrm{~m}$
C) width $7 \mathrm{~m}$, length $19 \mathrm{~m}$
D) width $8 \mathrm{~m}$, length $13 \mathrm{~m}$

Question

Quizplus · Accepted Answer

The original perimeter is $2l + 2w = 52$. The modified perimeter is $2(l + 24) + 2(2w) = 114$. Solving these equations, we find $l = 19$ and $w = 7$.

Solve the Problem 52 m52 \mathrm{~m}52 m If the Width Were Doubled and the Length Were

Solve the Problem $52 \mathrm{~m}$ If the Width Were Doubled and the Length Were