The perimeter of the rectangle is 70 m, its length is 1 m more than the width of which the sides of this rectangle are equal.

Let’s denote by x the length of this rectangle, and by y – its width.
According to the condition of the problem, the perimeter of this rectangle is 70 m, therefore, the following relationship takes place:
2 * (x + y) = 70.
It is also known that the length of this rectangle is 1 m greater than its width, therefore, the following relationship holds:
x = 1 + y.
We solve the resulting system of equations.
Substituting into the first equation the value x = 5 + y from the second equation, we get:
2 * (1 + y + y) = 70;
2 * y + 1 = 70/2;
2 * y + 1 = 35;
2 * y = 35 – 1;
2 * y = 34;
y = 24/2;
y = 17 m.
Knowing y, we find x:
x = 1 + y = 1 + 17 = 18 m.

Answer: The sides of this rectangle are 18 m and 17 m.