1. **State the problem:** We are given a system of equations: $$ \begin{cases} x_1 + x_2 + s_1 = 29 \\ 2x_1 + x_2 + s_2 = 30 \\ 4x_1 + x_2 + s_3 = 44 \end{cases} $$ and a partially completed table of basic solutions. We need to find the basic solution labeled (I). 2. **Recall the concept of basic solutions:** A basic solution corresponds to setting some variables to zero and solving for the others. The system has 5 variables ($x_1, x_2, s_1, s_2, s_3$) and 3 equations, so each basic solution sets exactly 2 variables to zero and solves for the remaining 3. 3. **Analyze the table:** The table shows values of variables for solutions (A) through (J). For (I), the last two columns $s_3$ and one other variable are zero. We need to find which variables are zero in (I) and solve accordingly. 4. **Identify zero variables for (I):** From the table, (I) has zeros in the last two columns, so $s_3=0$ and $s_2=0$ (assuming the order $s_1, s_2, s_3$). Thus, for (I): $$s_2 = 0, \quad s_3 = 0$$ 5. **Solve the system with $s_2=0$ and $s_3=0$: ** The system becomes: $$ \begin{cases} x_1 + x_2 + s_1 = 29 \\ 2x_1 + x_2 = 30 \\ 4x_1 + x_2 = 44 \end{cases} $$ 6. **Subtract the second equation from the third:** $$ (4x_1 + x_2) - (2x_1 + x_2) = 44 - 30 \\ 2x_1 = 14 \\ x_1 = 7 $$ 7. **Substitute $x_1=7$ into the second equation:** $$ 2(7) + x_2 = 30 \\ 14 + x_2 = 30 \\ x_2 = 16 $$ 8. **Find $s_1$ from the first equation:** $$ 7 + 16 + s_1 = 29 \\ s_1 = 29 - 23 = 6 $$ 9. **Write the basic solution (I):** $$ (x_1, x_2, s_1, s_2, s_3) = (7, 16, 6, 0, 0) $$ This satisfies the system and the zero conditions for $s_2$ and $s_3$. **Final answer:** Basic solution (I) is $(7, 16, 6, 0, 0)$.