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 150955: 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 nerdybill, josmiceli:
Answer by nerdybill(7384) (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, and its area is 192 square inches. What are the dimensions?
.
Let L = length of the rectangle
and W = width of the rectangle
.
For a 'rectangle' we know the following:
perimeter = 2(L+W)
area = LW
.
This then, allows us to derive our "system of equations":
56 = 2(L+W) (equation 1)
192 = LW (equation 2)
.
Solving equation 2 for W we get:
192 = LW
192/L = W
.
Substitute the above into equation 1 and solve for L:
56 = 2(L+W)
56 = 2(L + 192/L)
56L = 2(L^2 + 192)
28L = (L^2 + 192)
0 = L^2 - 28L + 192
Factoring:
0 = (L-16)(L-12)
.
L = {16,12}
Therefore, the dimensions are:
12 inches by 16 inches

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
in2
in
Let = length
Let = width

multiply both sides by

divide both sides by

Solve by completing the square

Take the square root of both sides

I'll graph it to check

Looks like the length can be either or
If ,

The length is 16 and the width is 12 (answer)
If

This is really the same answer with
which it shouldn't be
check answer:

OK