SOLUTION: perimeter of a rectangle is 22 inches and its length is 4 inches less than twice its width. Set up a system of linear equations and solve to find the dimensions of the rectangle.

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: perimeter of a rectangle is 22 inches and its length is 4 inches less than twice its width. Set up a system of linear equations and solve to find the dimensions of the rectangle.      Log On

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Question 4714: perimeter of a rectangle is 22 inches and its length is 4 inches less than twice its width. Set up a system of linear equations and solve to find the dimensions of the rectangle.
Found 2 solutions by xcentaur, rapaljer:
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
P=2(l+b)=22" ---[1]
l=b-4 ----------[2]
2(l+b)=22
(l+b)=11
given l=b-4
(b-4)+b=11
2b-4=11
2b=11+4
b=15/2
=7.5"
l=b-4=7.5-4=3.5"


ANS: 3.5"*7.5"


Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
I think there is an error in the previously posted solution.

Let W = Width of the rectangle
L = Length of the rectangle

Perimeter: 2W + 2L = 22
2nd Equation: L = 2W - 4

Solve by substituting L from the second equation into the first equation:
2W + 2(_____) = 22
2W + 2(2W -4) = 22
2W + 4W - 8 = 22
2W = 30
W= 5

Substitute into second equation:
L = 2W - 4
L = 2(5) - 4
L = 10 - 4 = 6

Check:
P = 2W + 2L
P = 2(5) + 2(6)
P = 10 + 12 = 22
AND
L = 2W - 4
6 = 2(5) - 4
6 = 10 - 4

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