SOLUTION: The perimeter of a rectangle is 56 cm. The length is 8 cm less than twice the width. Find the length and the width.

Question 450624: The perimeter of a rectangle is 56 cm. The length is 8 cm less than twice the width. Find the length and the width.
Answer by blwinbbbles(106) (Show Source): You can put this solution on YOUR website!
l = length w = width
Start with what you know and form equations. The perimeter is the distance of all sides added..length + length + width + width is equal to the perimeter or 2(length) + 2 (width) = perimeter
So: you could say 2l + 2w = 56 then you know that the length is 8 less then twice the width. l = 2w - 8..now that we have 2 equations we can find the solutions.
Since we know that l = 2w - 8 we can substite l from this equation into the other one.
2(2w - 8) + 2w = 56
4w - 16 + 2w = 56
6w - 16 = 56 add 16 to both sides
6w = 72 divide both sides by 6
w = 12
Now we know that w = 12...so plug that back into an equation
2l + 2(12) = 56
2l + 24 = 56 subtract 24 from both sides
2l = 32 divide both sides by 2
l = 16
Make sure these numbers fit both equations..so now you know the
Width of the rectangle is 12 cm and the length is 16 cm.
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