SOLUTION: Use a system of equations to solve the word problem. The perimeter of a rectangle is 56 inches, and its area is 192 square inches. What are the dimensions?

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Question 150954: Use a system of equations to solve the word problem. The perimeter of a rectangle is 56 inches, and its area is 192 square inches. What are the dimensions?
Found 2 solutions by stanbon, checkley77:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Use a system of equations to solve the word problem.
The perimeter of a rectangle is 56 inches
2L + 2W = 56
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and its area is 192 square inches
L*W = 192
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What are the dimensions?
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Rearrange to get:
L = 28 -W
Substitute to get:
(28-W)W = 192
28w - W^2 = 192
W^2 -28W + 192 = 0
(w-12)(w-16) = 0
W = 12 and L = 16
or
W = 16 and L = 12
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Cheers,
Stan H.

Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website!
2x+2y=56
xy=192
x=192/y
2(192/y)+2y=56
384/y+2y=56
(384+2y^2)/y=56
2y^2+384=56y
2y^2-56y+384=0
2(y^2-28y+192)=0
2(y-16)(y-12)=0
y-16=0
y=16 answer.
2x+2*16=56
2x+32=56
2x=56-32
2x=24
x=24/2
x=12 answer.
poof:
12*16=192
192=192