The length of a rectangle is 3 in. more than twice its width. If the perimeter of the rectangle is 18 in., what are the width of the rectangle?

We'll put the length of the rectangle to be a inches and the
width be b inches.

We know, from enunciation, that the length is 3
inches more than twice its width and we'll write the constraint
mathematically:

a - 3 = 2b

We'll
subtract 2b and add 3 both sides:

a - 2b = 3
(1)

The perimeter of the rectangle is 18
inches.

We'll write the perimeter of the
rectangle:

P = 2(a+b)

18 =
2(a+b)

We'll divide by 2:

9 = a +
b

We'll use the symmetric property:

a +
b = 9 (2)

We'll add (1) + 2*(2):

a - 2b
+ 2a + 2b = 3 + 18

We'l eliminate and combine like
terms:

3a = 21

We'll divide by
3:

a = 7 inches

7 + b =
9

b = 9 - 7

b  =  2
inches

So, the length of the rectangle is of 7 inches
and the width of the rectangle is of 2 inches.