The length and width of the rectangle is 8: 3. Find the length and width

The length and width of the rectangle is 8: 3. Find the length and width of the rectangle if it is known that its length is 40 dm greater than its width.

Let’s denote by x the length of this rectangle, and by y – its width.
According to the condition of the problem, the length and width of this rectangle are related as 8: 3, therefore, the following ratio takes place:
x / y = 8/3.
It is also known that the length of this rectangle is 40 dm larger than its width, therefore, the following relationship holds:
x = y + 40.
We solve the resulting system of equations.
Substituting into the second equation the value x = (8/3) * y from the first equation, we get:
(8/3) * y = y + 40.
We solve the resulting equation:
(8/3) * y – y = 40;
(5/3) * y = 40;
y = 40 / (5/3);
y = 40 * (3/5);
y = 3 * 40/5;
y = 3 * 8;
y = 24 dm.
Knowing y, we find x:
x = y + 40 = 24 + 40 = 64 dm.

Answer: the length of this rectangle is 64 dm, its width is 24 dm.