SOLUTION: Suppose that the width of a certain rectangle is 3 inches more than one-fourth of its length. The perimeter of the rectangle is 36 inches. Find the length and width of the rectangl

Question 765274: Suppose that the width of a certain rectangle is 3 inches more than one-fourth of its length. The perimeter of the rectangle is 36 inches. Find the length and width of the rectangle.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let the length of the rectangle = x
the width of the rectangle will be equal to x/4 + 3
the perimeter of the rectangle is equal to 36.
perimet of the rectangle is equal to 2 * length + 2 * width.
perimeter of the rectangle is equal to 2 * x + 2 * (x/4 + 3)
use the distributive law to get:
perimeter = 2 * x + 2 * (x/4) + (2 * 3)
simplify to get:
perimeter = 2x + x/2 + 6
simplify to get:
perimeter = 5x/2 + 6
since perimeter = 36, this formula becomes:
36 = 5x/2 + 6
subtract 6 from both sides of this equation to get:
30 = 5x/2
multiply both sides of this equation by 2 to get:
60 = 5x
divide both sides of this equation by 5 to get:
12 = x
your answer should be that x = 12.
let's see if that's accurate.
the length of the table is equal to 12
the width of the table is equal to 12/4 + 3 = 6
perimeter equals 2 * length + 2 * width = 2 * 12 + 2 * 6 = 24 + 12 = 36
answer is confirmed to be good.
the length of the table is equal to 12.
the width of the table is equal to 6.

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