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| Question 160339:  The area of a rectangle is 30 cm^2. The perimeter is 26cm. What are the length and width of the rectangle.
 this is what i have so far
 a=lw so 30= lw
 p=2L + 2W  so 26= 2L+2W
 I'm not sure if I am supposed to substitute the first equation into the second one or just try to isolate L and w.
 thanks.
 
 Found 2 solutions by  checkley77, Electrified_Levi:
 Answer by checkley77(12844)
      (Show Source): 
You can put this solution on YOUR website! xy=30 or x=30/y 2x+2y=26
 Now substitute (30/y) for x in the second equation & solve for y'
 2(30/y)+2y=26
 60/y+2y=26
 (60+2y*y)/y=26
 (60+2y^2)/y=26 now cross multiply;
 60+2y^2=26y
 2y^2-26y+60=0
 2(y^2-13y+30)=0
 2(y-10)(y-3)=0
 y-10=0
 y=10 answer.
 y-3=0
 y=3 answer.
 Proof:
 3*10=30
 30=30
 2*10+2*3=26
 20+6=26
 26=26
 
 
Answer by Electrified_Levi(103)
      (Show Source): 
You can put this solution on YOUR website! Hi, Hope I can help, .
 The area of a rectangle is 30 cm^2. The perimeter is 26cm. What are the length and width of the rectangle.
 this is what i have so far
 a=lw so 30= lw
 p=2L + 2W so 26= 2L+2W
 I'm not sure if I am supposed to substitute the first equation into the second one or just try to isolate L and w.
 .
 You did everything right so far.
 .
 You have the two equations
 .
 
  .
 
  .
 (You can reduce the second equation by dividing each side by "2",
  =  =  =  ) .
 The two equations will =
 .
 
  .
 
  .
 We have two equations, or a system of equations
 .
 To find the "L" and "W", we will use the way I solve systems of equations, which is pretty easy
 .
 
  .
 
  .
 First solve for a letter ( doesn't matter which one, usually the easiest), we will solve for "W" in both equations
 .
 First equation
  .
 To get "W" by itself we will divide each side by "L"
 .
 
  =  =  , Our first answer =  .
 Second equation
  .
 To get "W" by itself we will move "L" to the right side
 .
 
  =  =  or  .
 Our second answer =
  .
 We can put our two answers in an equation, since both answers represent "W"(width), our two answers equal each other
 .
 
  =  .
 We will use cross multiplication
 .
 
  =  =  .
 It becomes
  =  .
 This is a quadratic equation, we will put it in standard form(move "30" to the left side)
 .
 
  =  =  .
 Standard form
  Since this is a quadratic equation, we can use quadratic formula( We can solve by factoring, or completing the square as well, but I believe the formula is easiest) .
 
  ,  , (a = (-1), b = 13, c = (-30) quadratic formula =
  , replace the letters with numbers .
 
  =  =  =  = .
 
  =  =  .
 
  =  =  .
 "L" can be either
  or  .
 Just replace "L" in the second equation
 .
 
  =  =  , or .
 
  =  =  .
 Most likely
  and  , it could be the other way around, you can check by replacing these answers in the first two original equations. .
 Hope I helped, Levi
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