SOLUTION: Suppose that the length of a certain rectangle is 9 meters less than two times its width. The perimeter of the rectangle is 42 meters. Find the length and width of the rectangle.

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Question 213480: Suppose that the length of a certain rectangle is 9 meters less than two times its width. The perimeter of the rectangle is 42 meters. Find the length and width of the rectangle.
Answer by drj(1380) About Me  (Show Source):
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Suppose that the length of a certain rectangle is 9 meters less than two times its width. The perimeter of the rectangle is 42 meters. Find the length and width of the rectangle.

Step 1. Let l = 2w-9 be the length of rectangle and let w be the width

Step 2. Let P = 42 meters be the perimeter. Perimeter means adding the 4 sides of a rectangle. So,

P=2w-9%2B2w-9%2Bw%2Bw

P=6w-18

P=6w-18=42

Step 3. Add 18 to both sides of equation to get 4w by itself


P=6w%2B18-18=42%2B18

6w=60

Step 4. Divide 6 to both sides of equation

6w%2F6=60%2F6

w=+10+

Step 5. w = 10 meters is the width of the rectangle and length is 11 m since

l=2w-9=11

Check P=2w+2l=2(10)+2(11)=20+22=42 So w = 10 meters and l = 11 meters is the solution.

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Respectfully
Dr J