SOLUTION: Set up an equation and solve the problem. Suppose that the width of a certain rectangle is 6 inches less than its length. The area is numerically 16 less than twice the perimeter

Question 346560: Set up an equation and solve the problem.
Suppose that the width of a certain rectangle is 6 inches less than its length. The area is numerically 16 less than twice the perimeter. Find the length and width of the rectangle.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Set up an equation and solve the problem.
Suppose that the width of a certain rectangle is 6 inches less than its length.
The area is numerically 16 less than twice the perimeter.
Find the length and width of the rectangle.
:
Let x = the width
then
(x+6) = the length
And the area
A = x(x+6)
A = x^2 + 6x
and the perimeter
P = 2(x+6) + 2x
P = 2x + 12 + 2x
P = 4x + 12
:
"The area is numerically 16 less than twice the perimeter."
x^2 + 6x = 2(4x + 12) - 16
x^2 + 6x = 8x + 24 - 16
x^2 + 6x = 8x + 8
Arrange as a quadratic equation on the left:
x^2 + 6x - 8x - 8 = 0
x^2 - 2x - 8 = 0
Factors to
(x-4)(x+2) = 0
Positive solution is what we want here
x = 4 inches is the width
then
4 + 6 = 10 inches is the length
:
:
Check solution
Find the perimeter
2(10) + 2(4) = 28 in
Find the area
10*4 = 40 sq/in
:
Check in the statement:
"The area is numerically 16 less than twice the perimeter."
40 = 2(28) - 16
40 = 56 - 16; confirms our solution of x = 4

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