# The length of the rectangle is 12 dm longer than the width. If the length is increased by 3 dm and the width by 2 dm

**The length of the rectangle is 12 dm longer than the width. If the length is increased by 3 dm and the width by 2 dm, then the area of the rectangle will increase by 80 dm2. Find the length and width of the rectangle.**

Let x dm be the width of the rectangle, then the length of the rectangle is (x + 12) dm. The area of a rectangle is equal to the product of length and width. So S = x * (x + 12)

If the length is increased by 3 dm, then we get (x + 12 + 3) dm, that is, (x + 15) dm. If the width is increased by 2 dm, then we get (x + 2) dm. The area has increased by 80 dm ^ 2, which means the area is x (x + 12) + 80. So,

(x + 15) (x + 2) = x (x + 12) + 80

x ^ 2 + 15x + 2x + 30 = x ^ 2 + 12x + 80

x ^ 2 – x ^ 2 + 17x – 12x = 80 – 30

5x = 50

x = 10

The width is 10 dm, which means the length is 10 + 12 = 22 dm.

Answer: 10 dm and 22 dm