**A swimming pool is 8 ft longer than it is wide. The pool is surrounded by a walkway of width 4 ft. The combined area of the pool and the walkway is 1280 ft^2.**

**Answer:**

Obviously, we need to consider a variable for the width of the pool, since we are not given an actual number. For this reason, let the width of the pool be x. Notice that I am suppressing the (ft) right now.

Width of Pool (in feet) =x

So, now we can say that the length of the pool is x+8, since we were told that the swimming pool is 8 ft longer than it is wide.

Now, we are given a combined area of 1280-ft^2, when a 4-ft-wide walkway is added. So, consider what that gives us:

If the width without the walkway is $x, then the width of the pool surrounded by walkway must be x+4+4. This is because 4-ft was added to both sides of the pool.

By the same argument, since the length without the walkway was x+8, we can now say the length with the walkway is x+8+4+4. This is because 4-ft was added to the other two sides of the pool.

Now, we know that the pool with the walkway is x+8 feet wide and x+16 feet long. Since the area is 1280-ft^2, and area is:

length×width

we can now say:

1280=(x+16)(x+8)

Now, even though this doesn’t appear to show us much, we can isolate our x variable. This is useful, because x was the original width of the pool without the walkway. So, FOIL the right side of the equation to get a new polynomial on the right. Then, subtract 1280: