help me find the area for this problem please

Question

Suppose that the perimeter of a rectangle is 56 feet, and the length is 4 feet more than the width. Find the width of the rectangle, in feet.

Answer 1

First, remember that perimeter is equal to the distance around a rectangle. That means that the perimeter (p) is equal to 2 times the width (w) plus 2 times the length (l). so an equation for perimeter is

p= 2l + 2w

For this problem, we know that numerical value of the perimeter:

56 = 2l + 2w

Since we know that the length is 4 more than twice the width, we can represent the length l with this expression: 4 + 2w

Now our perimeter equation becomes:

56 = 2 (4 + 2w) + 2w

Using the distributive property, we can write

56 = 8 + 4w + 2w

Combining like terms gives us:

56 = 8 +6w

We can solve this equation using inverse operations (i.e. subtract 8 from both sides , then divided both sides of the equation by 6):

56 - 8 = 8 + 6w - 8

48 = 6w

48/6 = 6w/6

8 feet = w