The length of a rectangle is two units more than four times the width. What is the length if the perimeter is of 20 units?

We have to specify that the length of a rectangle is
bigger than the width.

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

We know, from
enunciation, that the width is 2 units more than 4 times it's length and we'll write the
constraint mathematically:

a - 2 =
4b

We'll subtract 4b and add 2 both
sides:

a - 4b = 2 (1)

The
perimeter of the rectangle is 20 units.

We'll write the
perimeter of the rectangle:

P =
2(a+b)

20 = 2(a+b)

We'll
divide by 2:

10 = a + b

We'll
use the symmetric property:

a + b = 10
(2)

We'll add (1) + 4*(2):

a -
4b + 4a + 4b = 2 + 40

We'l eliminate and combine like
terms:

5a = 42

We'll divide by
5:

a = 42/5 

a = 8.4
units

8.4 + b = 10

b = 10 -
8.4

b  =  1.6
units

So, the width of the rectangle is of
8.4 units and the lengths of the rectangle is of 1.6
units.

Since the width cannot
be larger than the length, we'll change and we'll put the length of 8.4 units and the
width of 1.6 units.