1. **State the problem:** We need to find the values of voltages $U_1$, $U_2$, $U_3$, and $U_4$ given the system of equations: $$ \begin{cases} U_1 + U_2 + U_3 = 220 \\ U_1 + U_4 = 200 \\ U_2 + U_3 = 140 \\ U_3 + U_4 = 160 \end{cases} $$ 2. **Use substitution and elimination:** From the second equation, express $U_1$ as: $$U_1 = 200 - U_4$$ 3. Substitute $U_1$ into the first equation: $$ (200 - U_4) + U_2 + U_3 = 220 $$ Simplify: $$ 200 - U_4 + U_2 + U_3 = 220 $$ $$ U_2 + U_3 - U_4 = 20 $$ 4. From the third equation, we have: $$ U_2 + U_3 = 140 $$ Subtract the equation from step 3: $$ (U_2 + U_3) - (U_2 + U_3 - U_4) = 140 - 20 $$ $$ \cancel{U_2} + \cancel{U_3} - \cancel{U_2} - \cancel{U_3} + U_4 = 120 $$ $$ U_4 = 120 $$ 5. Substitute $U_4 = 120$ into the second equation: $$ U_1 + 120 = 200 $$ $$ U_1 = 200 - 120 = 80 $$ 6. Substitute $U_4 = 120$ into the fourth equation: $$ U_3 + 120 = 160 $$ $$ U_3 = 160 - 120 = 40 $$ 7. Substitute $U_3 = 40$ into the third equation: $$ U_2 + 40 = 140 $$ $$ U_2 = 140 - 40 = 100 $$ **Final answer:** $$ U_1 = 80, \quad U_2 = 100, \quad U_3 = 40, \quad U_4 = 120 $$