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 ?
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.