206. Find the number which, when increased by two thirds of itself and one unit, gives 10.

Note. The last two problems are solved by the method of inversion or, alternatively, the method of reverse actions.
206. $x+\frac{2}{3} x+1=10 ; \frac{5}{3} x=9 ; x=5 \frac{2}{5}$.

But Bega Ed-Din solves it by the "rule of the balance" (regula falsi), found in the works of ibn-Ezra, ibn-Albanna, and Al-Kalsadi. The essence of the method is as follows: let the equation be $a x+b=0$. Substitute arbitrary values $z_{1}$ and $z_{2}$ for $x$. We will have $a z_{1}+b=k_{1}$ and $a z_{2}+b=k_{2}$. The quantities $k_{1}$ and $k_{2}$ are called the errors of the equation. It is easy to see that

$$
\begin{aligned}
& a\left(x-z_{1}\right)=-k_{1} \\
& a\left(x-z_{2}\right)=-k_{2}
\end{aligned}
$$

thus: $\frac{k_{1}}{x-z_{1}}=\frac{k_{2}}{x-z_{2}}$, from which $\frac{x-z_{1}}{x-z_{2}}=\frac{k_{1}}{k_{2}}$.

The differences $x-z_{1}$ and $x-z_{2}$ are called the errors of the substitutions. From the last equality:

$$
x=\frac{k_{9} z_{1}-k_{1} z_{2}}{k_{2}-k_{1}}
$$

Bega Ed-Din assumes $z_{1}=9$. Then $9+\frac{2}{3} \cdot 9+1=16$ (error $k_{1}=6$); then $z_{2}=6 ; 6+\frac{2}{3} \cdot 6+1=11$; the second error $k_{2}=1$; then:

$$
x=\frac{1 \cdot 9-6 \cdot 6}{1-6}=\frac{27}{5}
$$