SOLUTION: The perimeter of rectangle is 310 m. The length is 25 m greater than the width. Find the width and the length of the rectangle. I do not know how to actually solve this usin

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The perimeter of rectangle is 310 m. The length is 25 m greater than the width. Find the width and the length of the rectangle. I do not know how to actually solve this usin      Log On


   

Question 8396: The perimeter of rectangle is 310 m. The length is 25 m greater than the width. Find the width and the length of the rectangle.
I do not know how to actually solve this using an equation. I am using a "guess and check" method. The answer I have is 90 length and 65 width, it all adds up, but I don't know how to show it.
PLEASE HELP
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, your guess-and-check method gave you the right answer.
Here's how you might do it algebraically:
P = 2(L+W)= 310 m The formula for the perimeter of a rectangle.
L = W+25 m The length is 25 m greater than the width.
P = 2((W+25) + W) = 310 Simplify and solve for W
2(2W + 25) = 310
4W + 50 = 310 Subtract 50 from both sides.
4W = 260 Divide both sides by 4.
W = 65 m and L = W + 25 = 65 + 25 = 90 m.
So the length is 90 m and the width is 65 m.