Question:The perimeter of a rectangle is 44 m . The length is 2 m greater than the width. Find the length and width  of the rectangle.. 

Answer The perimeter of the rectangle is x 2(x+y) meters.
                                                 
                                                  `:.` 2(x+y) =44.

Divide both sides of this equations by 2;
                                            
                                              x+y = 22. . . . . . . . . . . . . . . . . . . . .(1) 


We are also told that the length is 2 m greater than the width.

                                           `:.` x = y+2, 
                                                x - y = 2 . . . . . . . . . . .  . . . . . . . . .(2)

we have the two equations 
                             
                                      x + y = 22 . . . . . . . . . . . . .(1)
                                      x - y  = 2  . . . . . . . . . . . . . (2)

add:                                   2x = 24.
                 
                              `:.`  x = 12.

Substitute in (1) :         12 + y = 22
                
                             `:.` y = 10.

The length of the rectangle is 12 m ; its width is 10  m. 

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