Solve the problem.
-The perimeter of a rectangle is $78 \mathrm {~m}$. If the width were doubled and the length were increased by $12 \mathrm {~m}$, the perimeter would be $132 \mathrm {~m}$. 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 $15 \mathrm {~m}$, length $24 \mathrm {~m}$
D) width $24 \mathrm {~m}$, length $15 \mathrm {~m}$

Question

Quizplus · Accepted Answer

The original perimeter of the rectangle is given by $2l + 2w = 78$, where $l$ is the length and $w$ is the width. When the width is doubled and the length is increased by 12 m, the new perimeter becomes $2(l + 12) + 2(2w) = 132$. Solving these equations simultaneously gives the length $l = 24 \mathrm {~m}$ and the width $w = 15 \mathrm {~m}$.

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