One of the sides of the rectangle was reduced by 20% and the other was increased by x

One of the sides of the rectangle was reduced by 20% and the other was increased by x; it turned out that the rectangle and the area did not change, find x.

Let a rectangle be given with width a, length b, then its perimeter will be equal to 2 ∙ (a + b), and the area will be equal to (a ∙ b). The condition of the problem says that one of the sides of the rectangle was reduced by 20%, that is, it became (0.8 ∙ a). It is known that the other side of the rectangle was increased by x, that is, it became (b + x). At the same time, its perimeter became equal to 2 ∙ (0.8 ∙ a + b + x), and the area became equal to 0.8 ∙ a ∙ (b + x). Knowing that after the changes it turned out that the perimeter and area did not change, we compose a system of equations:
2 ∙ (a + b) = 2 ∙ (0.8 ∙ a + b + x) and a ∙ b = 0.8 ∙ a ∙ (b + x).
After opening the brackets and bringing similar terms, we get, a = 5 ∙ x, b = 4 ∙ x, that is, a: b = 5: 4.
Let us introduce the coefficient of proportionality k, then a = 5 ∙ k and b = 4 ∙ k. Let’s substitute these expressions into the system:
2 ∙ (5 ∙ k + 4 ∙ k) = 2 ∙ (0.8 ∙ 5 ∙ k + 4 ∙ k + x) and 5 ∙ k ∙ 4 ∙ k = 0.8 ∙ 5 ∙ k ∙ (4 ∙ k + x).
After expanding the brackets and reducing similar terms, we find that k = 1 and x = 1.
Answer: unknown term x = 1.