Solve the Problem $60 \mathrm {~m}$ If the Width Were Doubled and the Length Were

Multiple Choice

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: $21 \mathrm {~m}$