SOLUTION: The perimeter of a rectangle is 40m. If the width is increased by twice the length, the result is 32m. Find the length and width of the rectangle.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The perimeter of a rectangle is 40m. If the width is increased by twice the length, the result is 32m. Find the length and width of the rectangle.      Log On


   

Question 386816: The perimeter of a rectangle is 40m. If the width is increased by twice the length, the result is 32m. Find the length and width of the rectangle.
Answer by gwendolyn(128) About Me  (Show Source):
You can put this solution on YOUR website!
Let L be the length of the rectangle
Let W be the width of the rectangle

The perimeter of a rectangle is 40m.
So:
2*L + 2*W = 40

If the width is increased by twice the length, the result is 32m.
So:
W + 2L = 32
Solve for W by subtracting 2L from both sides:
W + 2L - 2L = 32 - 2L
W = 32 - 2L

Substitute this value of W back into the first equation:
2*L + 2*W = 40
2*L + 2*(32 - 2L) = 40
Distribute the 2:
2L + 2*32 - 2*2L = 40
2L + 64 - 4L = 40
Collect the L terms:
-2L + 64 = 40
Subtract 64 from both sides to isolate the L term:
-2L + 64 - 64 = 40 - 64
-2L = -24
Divide both sides to solve for L
(-2L)/(-2) = (-24)/(-2)
L = 12

Substituting the value of L into our second equation:
W = 32 - 2L
W = 32 - 2*12
W = 32 - 24
W = 8