# 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?

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

Given that the length is 3 more than twice the width:

==> we will write"

L= 3 + 2W..............(1)

Also, given the perimeter is :

P = 18 in

But we know that the perimeter is: 2L + 2W

==> 2L + 2W = 18

==> L + W = 9

But L = 3+ 2W

==> ( 3+ 2W) + W = 9

==> 3+ 3W = 9

==> 3W = 6

==> W= 2

==> L = 7

**Then the length of the rectangle = 7 in**

**and the width of the rectangle = 2 in**

Let the width of the rectangle be x.Then the length is 3 in more than twice width implies length = 2x+3 inches.

Therefore the perimeter of the rectangle =2 (width+length) = 2*(x+2x+3) which is given to be 18 in.

So 2(x+2x+3) = 18. Now we solve for x.

2(3x+3) = 18.

6(x+1) = 18.

x+1 = 18/6 = 3.

x= 3-1 = 2

x = 2.

So the width of the rectangle = 2 and the length = 2x+3 = 4+3 = 7.

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.**