The length of a rectangular picture frame is 2 inches longer than twice the width. if the perimeter of the frame is 34 inches, find the dimensions of the frame.

The length, l, is 2 more than twice the width, w.
l= 2w+2 (2 more than twice w)
In a rectangle, we have two of the same lengths and two of the same widths.
2l+2w=34
Divide by 2 on each side.
l+w=17
We remember that l is 2w+2
Substituting that back in we get,
2w+2+w=17
3w+2=17
3w=15
w=5
We sub w back into l+w=17
l+5=17
l=12
The dimensions are 5 by 12 inches.

Let,
the length of the rectangular picture frame be "x"
the width of the rectangular picture frame be "y"
Then, according to the question, 
x = 2 + 2y equation (1)

Perimeter = 34 inches

Now,
We have,
perimeter of the rectangular = 2 ( length + width)
                                    34 = 2 ( x + y)
                                    34 = 2 ( 2 + 2y + y)
                                    34 = 2 (2 + 3y)
                                    34 = 4 + 6y
                               34 - 4 = 6y
                                    30 = 6y
                               30 / 6 = y
                                      5 = y
Now,
taking equation (1)
x = 2 + 2y
substituting the value of "y" , we get,
x = 2 + 2(5)
x = 2 + 10
x = 12

So, the dimensions of the rectangular frame are 12 inches and 5 inches