PLEASE HELP......JUST SHOW THE WORK SO I CAN UNDERSTAND!!!! ?

Question

A swimming pool is 5 m longer than it is wide and is surrounded by a deck 2 m wide. The area of the pool and deck togetha is at least 140 m squared greater than the area of the pool alone. What can you conclude about the dimensions of the pool?

Answer 1

Let the pool's length be L.  Then we know the width is L - 5.  (The length is 5 m longer than the width.)  Make a sketch:  you should see that the dimensions of the deck are, then, L + 4 and (L - 5) + 4 = L - 1.  (It's four instead of two because the deck is going to be on both sides of the pool.)

We know that you get the area of rectangles by multiplying length times width.  The area of the pool will be L(L - 5).  The area of the deck and pool together will be (L + 4)(L - 1).  We know that this latter area is at least 140 more than the former, so

(L + 4)(L - 1) - L(L - 5) >= 140

"Greater than or equal to" because it's "at least" 140 greater.

I'll let you work it out from there.