# The length of a rectangle is 5 in. more than its width. If the perimeter of the rectangle is 30 in. what are the width of the rectangle ?

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### 1 Answer

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 5 inches more than twice its width and we'll write the constraint mathematically:

a - 5 = 2b

We'll subtract 2b and add 3 both sides:

a - 2b = 5 (1)

The perimeter of the rectangle is 30 inches.

We'll write the perimeter of the rectangle:

P = 2(a+b)

30 = 2(a+b)

We'll divide by 2:

15 = a + b

We'll use the symmetric property:

a + b = 15 (2)

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

a - 2b + 2a + 2b = 5 + 30

We'll eliminate and combine like terms:

3a = 35

We'll divide by 3:

a = 11.66 inches

11.66 + b = 15

b = 15 – 11.66

b = 3.33 inches

**Therefore, the length of the rectangle is of 11.66 inches and the width of the rectangle is of 3.33 inches.**