# rectangle problemThe 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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First, we set up two equations based on the information you have given.

L = 2w + 3

2L + 2W = 18

Now, we use the value for L in the first equation and substitute it into the second equation.

2 (2W + 3) + 2W = 18

4W + 6 + 2W = 18

6W = 12

W = 2

Now we use this value for W to determine L.

L = 2*2 + 3

L = 4 + 3

L = 7

**So, the length is 7 inches and the width is 2 inches.**

The length of a rectangle is 3 inch more than twice its width.

Let the width of the rectangle be W. The length of the rectangle is 3 + 2*W

For a rectangle with sides W and L the perimeter is equal to 2*(W + L)

Substituting the values in the formula, the perimeter is:

2*(W + 3 + 2*W). This is equal to 18

2*(W + 3 + 2*W) = 18

W + 3 + 2*W = 9

3W + 3 = 9

3W= 6

W = 2

The width of the rectangle is 2 inches.

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.