13.006. A tractor driver plowed three plots of land. The area of the first is $2 / 5$ of the area of all three plots, and the area of the second is to the area of the third as $3 / 2: 4 / 3$. How many hectares were there in all three plots if the third plot was 16 hectares less than the first?
Solution.

Let the areas of the plots be $x, y, x-16$ (ha). According to the condition, $x=\frac{2}{5}(x+y-16+x)$ and $y:(x-16)=3 / 2: 4 / 3$, from which $y=\frac{9}{8}(x-16)$. We have
$x=\frac{2}{5}\left(2 x-16+\frac{9}{8} x-18\right) \Leftrightarrow x=\frac{2}{5}\left(\frac{25 x}{8}-34\right), x=\frac{272}{5}$. Next, we find $y=\frac{9}{8}\left(\frac{272}{5}-16\right)=\frac{216}{5}$. The total area is $\frac{544}{5}+\frac{216}{5}-16=136$ (ha).

Answer: 136 ha.