Solve the problem.
-The perimeter of a rectangle is $78 \mathrm {~m}$. If the width were doubled and the length were increased by $16 \mathrm {~m}$, the perimeter would be $138 \mathrm {~m}$. What are the length and width of the rectangle?
A) width $19 \mathrm {~m}$, length $19 \mathrm {~m}$
B) width $14 \mathrm {~m}$, length $19 \mathrm {~m}$
C) width $25 \mathrm {~m}$, length $14 \mathrm {~m}$
D) width $14 \mathrm {~m}$, length $25 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The original perimeter is $2l + 2w = 78$. After changes, the new perimeter is $2(l + 16) + 2(2w) = 138$. Solving these equations gives $l = 25$ and $w = 14$.

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