$v_l-\text{prędkosˊcˊ łoˊdki}$  

$v_t-\text{prędkosˊcˊ tratwy i jednoczesˊnie prędkosˊcˊ wody}$  

$t-\text{czas w jaki łoˊdka pokonuje 20 km zgodnie z nurtem rzeki}$  

 

$\{\begin{array}{l}\left(v_l+v_t\right)t=20\\ \left(v_l-v_t\right)\cdot \frac{8}{20}\left(7-t\right)=8\\ v_t\left(t+\frac{8}{20}\left(7-t\right)\right)=12\end{array}$           

$\{\begin{array}{l}{v}_l=\frac{20-v_{t}t}{t}\\ \left(v_l-v_t\right)\cdot \frac{8}{20}\left(7-t\right)=8\\ t-\frac{8}{20}t+\frac{56}{20}=\frac{12}{v_t}\end{array}$               

$\{\begin{array}{l}{v}_l=\frac{20-v_{t}t}{t}\\ \left(v_l-v_t\right)\cdot \frac{8}{20}\left(7-t\right)=8\\ t=\frac{20}{12}\left(-\frac{56}{20}+\frac{12}{v_t}\right)\end{array}$     

$\{\begin{array}{l}{v}_l=\frac{20-v_{t}t}{t}\\ \left(v_l-v_t\right)\cdot \frac{8}{20}\left(7-t\right)=8\\ t=\frac{20}{v_t}-\frac{56}{12}\end{array}$     

$\{\begin{array}{l}\left(\left(\frac{20-v_{t}t}{t}\right)-v_t\right)\cdot \frac{8}{20}\left(7-t\right)=8\\ t=-\frac{14}{3}+\frac{20}{v_t}\end{array}$     

Wstawmy wyznaczone t do pierwszego równania.

$\left(\frac{20-v_t\left(-\frac{14}{3}+\frac{20}{v_t}\right)}{-\frac{14}{3}+\frac{20}{v_t}}-v_t\right)\cdot \frac{8}{20}\left(7+\frac{14}{3}-\frac{20}{v_t}\right)=8$      

$\left(\frac{20-v_t\left(-\frac{14}{3}+\frac{20}{v_t}\right)+\frac{14}{3}{v}_t-20}{-\frac{14}{3}+\frac{20}{v_t}}\right)\cdot \frac{8}{20}\left(7+\frac{14}{3}-\frac{20}{v_t}\right)=8$        

$\left(20-v_t\left(-\frac{14}{3}+\frac{20}{v_t}\right)+\frac{14}{3}{v}_t-20\right)\cdot \frac{2}{5}\left(\frac{35}{3}-\frac{20}{v_t}\right)=8\left(-\frac{14}{3}+\frac{20}{v_t}\right)$    

$\left(\frac{28}{3}{v}_t-20\right)\cdot \left(\frac{14}{3}-\frac{8}{v_t}\right)=-\frac{112}{3}+\frac{160}{v_t}$  

$\frac{392}{9}{v}_t-\frac{224}{3}-\frac{280}{3}+\frac{160}{v_t}+\frac{112}{3}-\frac{160}{v_t}=0$   

$\frac{392}{9}{v}_t-\frac{224}{3}+\frac{112}{3}-\frac{280}{3}=0$  

$\frac{392}{9}{v}_t=\frac{392}{3}$  

$v_t=3$    

Woda płynie z prędkością 3 km/h.