# 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?

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Let us represent the width of the rectangle by W. The length is equal to two units more than four times the length or 2 + 4*W.

Now the perimeter is given as 20.

The perimeter is also equal to 2*W + 2*L

So 2*W + 2*L =20

=>2W + 2(2 + 4W) = 20

=> 2W + 4 + 8W = 20

=> 10 W = 16

=> W = 16/10

The length is 2 + 4W = 2 + 4*16/10 = 8.4

**Therefore the required length is 8.4.**

If the width is x, then the length is 2 more than 4times width. So length is 4x+2.

The perimeter p is 2(length+width) = 2(4x+2+x) = 2(5x+2).

But the actual perimeter = 20.

=> 2(5x+2) = 20

=> (5x+2) = 20/2 = 10.

=> 5x+2 = 10.

=> 5x = 10-2 = 8

=>x = 8/5 = 1.6.

=> 4x+2 = 1.6*4+2 = 8.4 units.

So width = 1.6 units, and length = 8.4 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.**