SOLUTION: I don't understand this word problem. The problem says "The perimeter of a rectangle is 60 inches., and its area is 200 inches^2. Find the dimensions of the rectangle." Could you p

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I don't understand this word problem. The problem says "The perimeter of a rectangle is 60 inches., and its area is 200 inches^2. Find the dimensions of the rectangle." Could you p      Log On


   

Question 120198: I don't understand this word problem. The problem says "The perimeter of a rectangle is 60 inches., and its area is 200 inches^2. Find the dimensions of the rectangle." Could you please help me?
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
PERIMETER=2L+2W
AREA=L*W
60=2L+2W
200=L*W OR L=200/W NOW SUBSTITUTE (200/W) FOR L IN THE FIRST EQUATION & SOLVE FOR W
60=2(200/W)+2W
60=400/W+2W ADD THE 2 TERMS ON THE LEFT SIDE OF THE EQUATION BY FINDING A COMMON DENOMINATOR (W) THUS:
60=(400+2W*W)/W
60=(400+2W^2)/W NOW CROSS MULTIPLY
400+2W^2=60W
2W^2-60W+400=0
2(W^2-30W+200)=0
2(W-20)(W-10)=0
W-20=0
W=20 ANSWER FOR THE WIDTH & 10 FOR THE LENGTH.
OR W-10=0
W=10 FOR THE WIDTH & 20 FOR THE LENGTH.
PROOF
60=2*10+2*20
60=20+40
60=60
&
200=10*20
200=200