### 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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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.**

Let the length of the rectangle be L and the width be W.

The length of a rectangle is two units more than four times the width.

=> L = 2 + 4*W

=> W = (L - 2)/4

The perimeter is of 20 units

=> 2*L + 2*W = 20

=> L + W = 10

substitute W = (L - 2)/4

=> L + (L - 2)/4 = 10

=> 4L + L - 2 = 40

=> 5L = 42

=> L = 42/5

**The length of the rectangle is 42/5**

The length of a rectangle is two units more than four times the width. If the length of a rectangle is L and the width is W, the perimeter of the rectangle is 2*(L+W).

Here, the perimeter of the rectangle is 20 units.

If the width is taken as x, the length is 4x + 2.

2*(x + 4x + 2) = 20

2*(5x + 2) = 20

10x + 4 = 20

10x = 16

x = 1.6

The length of the rectangle is 1.6*4 + 2 = 8.4

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