Two rectangles have the same width. The length of one is 2 feet longer than the width. The length of the other is 4 feet longer than the width. The larger rectangle has 8 more square feet than the smaller. What is the width of the rectangles?

arealarge-8=areasmall

w(w+4)-8=w(w+2)
solve for w.

To solve this problem, let's break it down step by step.

Let's start by assuming the width of the rectangles is represented by the variable 'w'.

According to the given information, the length of one rectangle is 2 feet longer than the width. We can represent this as 'w + 2'.

Similarly, the length of the other rectangle is 4 feet longer than the width, which can be represented as 'w + 4'.

Next, we are told that the larger rectangle has 8 more square feet than the smaller one. To find the area of a rectangle, we multiply its length by its width.

So, the area of the larger rectangle can be expressed as (w + 4) * w, and the area of the smaller rectangle can be expressed as (w + 2) * w.

According to the problem, the larger rectangle has 8 more square feet than the smaller one. Therefore, we can write the equation:

(w + 4) * w = (w + 2) * w + 8

Now, let's solve this equation step by step:

w^2 + 4w = w^2 + 2w + 8 (distributing w to both terms)
w^2 - w^2 + 4w - 2w = 8 (subtracting w^2 and 2w from both sides)
2w = 8 (combining like terms)
w = 4 (dividing both sides by 2)

Therefore, the width of the rectangles is 4 feet.