SOLUTION: Hi there, I just can't get this type of problem so I need some help. If anyone would like to solve this, I would be very greatful.
The perimeter of a rectangle is 132 inches. The
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-> SOLUTION: Hi there, I just can't get this type of problem so I need some help. If anyone would like to solve this, I would be very greatful.
The perimeter of a rectangle is 132 inches. The
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Question 191730: Hi there, I just can't get this type of problem so I need some help. If anyone would like to solve this, I would be very greatful.
The perimeter of a rectangle is 132 inches. The length exceeds the width by 58 inches. Find the length and width in inches. Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website! let L and W be the length and width of rectangle respectively
:
P=132............eq 1
P=2L+2W..........eq 2
L=W+58...........eq 3
:
take P's value from eq 2 and L's value from eq 3 and plug them into eq 1
:
132=2(W+58)+2W
:
132=2W+116+2W
:
16=4W
:
W=4
:
L=W+58=4+58=62
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