Question 73092
<pre>
The length of a rectangle is 2 cm more than twice its width.  
If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.

What is asked in the problem?
 find the dimensions of the rectangle.

Given:
 The length of a rectangle is 2cm more than twice its width
 The perimeter of the rectangle is 52cm

Representation:
  Let x = the width of the rectangle
     2x+2 = the length of the rectangle

Equation
          2l + 2w = P
     2(2x+2) + 2x = 52        Multiply (Distributive Property)
      4x + 4 + 2x = 52        Combine like terms
           6x + 4 = 52
       6x + 4 - 4 = 52 - 4    subtract 4 both sides
               6x = 48        divide both sides by 6
                x = 8 cm -----> Width

     length = 2x + 2 
            = 2(8) + 2  
            = 16 + 2
            = 18 cm    ----- length

Checking

         2L + 2W = P
    2(18) + 2(8) = 52
         36 + 16 = 52
              52 = 52