How the Early Egyptians Solved Equation

$f(a) = a+1$ $f(ab) = ab + \frac{ab}{a}$ $=> f(ab) = ab+b$ $=> f(ab) = b(a+1)$ $=> f(ab) = b(f(a))$ $=>f(ab) = f(x)$ Since f(x) is one to one function $=>ab=x$

Encontremos $f(a)$ ,

$f(a) = a + \frac{a}{a} \\\\ f(a) = a + 1$

Ora, de acordo com o enunciado $f(x) = b(f(a))$ ,

$f(x) = b \cdot f(a) \\\\ b = \frac{f(x)}{f(a)} \\\\ b = \frac{x + \frac{x}{a}}{a + 1} \\\\ b = \frac{ax + x}{a} \cdot \frac{1}{a + 1} \\\\ b = \frac{x}{a} \\\\ \boxed{ab = x}$