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?
.
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