SOLUTION: I need help on this problem that I am simply stuck on.
The perimeter of a rectangle is 196 ft. The length of the rectangle is 40 ft. less than twice the width. Find the width an
Algebra.Com
Question 919411: I need help on this problem that I am simply stuck on.
The perimeter of a rectangle is 196 ft. The length of the rectangle is 40 ft. less than twice the width. Find the width and the length of the rectangle.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Let length = L, width = W
2(L+W) = 196 <--> L+W = 98
L = 2W - 40
Now you have two equations and two variables. Can you solve for L, W (hint: you can substitute L with 2W-40)?
RELATED QUESTIONS
Hi! I really need some help with this problem I've been stuck on.
The length of a... (answered by solver91311,Cromlix)
Hi! I really need help with this problem that I am currently stuck on. Could you help me? (answered by MathLover1,KMST)
The area of a rectangle is 52 ft sq., and the length of the rectangle is 5 ft less than... (answered by ankor@dixie-net.com,josmiceli)
I am working on a problem that I need some help on.
Here it is: Geometry. The length of... (answered by jim_thompson5910)
I am stuck on this problem. Please help. The perimeter of a rectangle is 34 feet and... (answered by tutorcecilia)
I would like help on how to solve this story problem: The length of a rectangle is 9ft... (answered by josgarithmetic)
I am stuck with where to begin... I think I need the perimeter of rectangle, area of... (answered by solver91311)
I have a word problem that I need to solve. I have set it up but I don't know if I did it (answered by josgarithmetic)
I am stuck on this problem, any help would be greatly appreciated.
Given a rectangle... (answered by rfer,lwsshak3)