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)   (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
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