SOLUTION: The perimeter of a rectangle is 84 feet. If the length was increased by 3 feet and the width was doubled, the perimeter would be 120 feet. What are the dimensions of the rectangle?
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Question 1077750: The perimeter of a rectangle is 84 feet. If the length was increased by 3 feet and the width was doubled, the perimeter would be 120 feet. What are the dimensions of the rectangle?
I spent hours on this problem please help! Found 2 solutions by Boreal, amfagge92:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! 2L+2W=84
L+W=42
L+3+2W=60
substitute L=42-W from above
42-W+3+2W=60
45-W+2W=60
W=15 feet
L=27 feet
L+3=30 feet
2W=30 feet
The first rectangle is 15 x 27, perimeter is 84
the second is a square, 30 x 30, perimeter is 120
You can put this solution on YOUR website! SOLUTION
PERIMETER OF RECTANGLE=2(l+b)
2(l+b)=84...i
2[(l+3)+2b)=120...ii
solving
l+b=42
l=42-b...iii
subtutute l in eqn ii
l+3+2b=60
42-b+3+2b=60
b=15
put b in eqn iii
l=42-15=27
Length =27, Breadth=15
check:2(l+b)=84
2(27+15)=84
