Solve the problem.
-The perimeter of a rectangle is $60 \mathrm {~m}$. If the width were doubled and the length were increased by 19 $\mathrm { m }$, the perimeter would be $116 \mathrm {~m}$. What are the length and width of the rectangle?
A) Length: $15 \mathrm {~m}$; width: $15 \mathrm {~m}$
B) Length: $21 \mathrm {~m}$; width: $9 \mathrm {~m}$
C) Length: $15 \mathrm {~m}$; width: $10 \mathrm {~m}$
D) Length: $9 \mathrm {~m}$; width: $21 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The original perimeter is $2l + 2w = 60$. After modifications, the new perimeter is $2(l + 19) + 4w = 116$. Solving these equations simultaneously gives the length $l = 21 \mathrm {~m}$ and width $w = 9 \mathrm {~m}$.

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