The length of the rectangle is 2 m longer than its width. If the width is increased by 3 m

The length of the rectangle is 2 m longer than its width. If the width is increased by 3 m and the length by 8 m, then the area will increase by 3 times. Find the sides of the rectangle .

1) Let x m be the width of the rectangle (x> 0), then (x + 2) m is its length.

2) (x * (x + 2)) = (x ^ 2 + 2x) m ^ 2 is the area of the rectangle.

3) (x + 3) m – the width of the rectangle after the enlargement, ((x + 2) + 8) = (x + 10) m – the length of the enlarged rectangle.

4) (x + 3) (x + 10) = (x ^ 2 + 13x + 30) m – the area of the enlarged rectangle.

5) From the condition, you can write:

(x ^ 2 + 13x + 30): (x ^ 2 + 2x) = 3.

6) Solve the equation:

x ^ 2 + 13x + 30 = 3 * (x ^ 2 + 2x);

x ^ 2 + 13x + 30 = 3x ^ 2 + 6x;

2x ^ 2 – 7x – 30 = 0.

D = 49 + 240 = 289;

x1 = -2.5, x2 = 6.

7) x1 <0, which means it is not a solution to the problem.

8) x = 6 m – the width of the rectangle.

9) 6 + 2 = 8 m – its length.

Answer: 8 and 6 m.