SOLUTION: the perimeter of a rectangle is 24 ft. Its length is five times its width. Let x be the length and y be the width. What is the area of the rectangle?

Question 266188: the perimeter of a rectangle is 24 ft. Its length is five times its width. Let x be the length and y be the width. What is the area of the rectangle?
Found 2 solutions by calegebra, MathTherapy:
Answer by calegebra(14) (Show Source): You can put this solution on YOUR website!

P=2L+2W Length (here)=5y Width (here)=y Substitute Length and Width (here) into the Perimeter equation as in: P=2L+2W = 24 = 2(5y)+(y)=10y+y=11y
If 11y=24 than y=24/11 that is width and x(length)=5y or 5(24/11)
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
the perimeter of a rectangle is 24 ft. Its length is five times its width. Let x be the length and y be the width. What is the area of the rectangle?

Since x = length and y = width, and since perimeter = 24, then 2L + 2W = P becomes 2x + 2y = 24

Also, since length or x is 5 times width or y, then we'll have: x = 5y

We can then substitute 5y for x in the 1st equation to get: 2(5y) + 2y = 24 ---- > 10y + 2y = 24

12y = 24

Since y or width = 2, then x or length = 5(2) = 10

Since width = 2 and length = 10, then area = L*W or 10 * 2 = feet squared.
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