A small rectangular garden has an area of 80 square feet. Its length is 2 feet more than the width. Give both the length and width of the garden.

Answer 1

This problem will include the solving of 2 algebraic equations. The Area of a rectangle is found by X * Y = A. We know that A = 80 so our first equation is: X * Y = 80 where Y is our length and X is our width.

Next, the problem tells us that the length (Y) is two feet more than the width. That means X (width) + 2 = Y (length). This is our next equation X+2 = Y.

Now we can substitute for Y in our first equation since we know the value of Y from our equation 2. By substituion, we know now that X * (X+2) = 80. From here, we use the distributive property to find X² +2x = 80. We move the 80 to the left side of the equation by subtracting 80 from each side to get us X²+2x-80 = 0.

We can factor this equation to (X-8)*(X+10) = 0 which gives us two solutions. These solutions for the Width are X = -10 ft and X = 8 ft. Since we are talking about a width of a garden, we can not have a negative width so we disregard X = -10. This leaves us with X = 8ft.

To solve for length, we plug X =8 into either of our equations. For this purpose, we will plug it into equation 2: X+2 = Y. 8+2 =Y so now we know length Y = 10 ft.