1. **State the problem:** We are given a system of equations: $$ \begin{cases} x_1 + x_2 + s_1 = 28 \\ 2x_1 + x_2 + s_2 = 34 \\ 3x_1 + x_2 + s_3 = 48 \end{cases} $$ and a table of basic solutions (A) through (J). We need to find the basic solution (I). 2. **Recall what a basic solution is:** 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. 3. **Analyze the pattern in the table:** Each solution sets two variables to zero and solves for the other three. 4. **Identify which variables are zero in solution (I):** From the table, solution (I) has zeros in the last two columns, which correspond to $s_2$ and $s_3$. 5. **Set $s_2 = 0$ and $s_3 = 0$ and solve the system:** $$ \begin{cases} x_1 + x_2 + s_1 = 28 \\ 2x_1 + x_2 + 0 = 34 \\ 3x_1 + x_2 + 0 = 48 \end{cases} $$ 6. **From the second and third equations:** $$ 2x_1 + x_2 = 34 \\ 3x_1 + x_2 = 48 $$ Subtract the second from the third: $$ (3x_1 + x_2) - (2x_1 + x_2) = 48 - 34 \\ x_1 = 14 $$ 7. **Substitute $x_1 = 14$ into the second equation:** $$ 2(14) + x_2 = 34 \\ 28 + x_2 = 34 \\ x_2 = 6 $$ 8. **Substitute $x_1 = 14$ and $x_2 = 6$ into the first equation:** $$ 14 + 6 + s_1 = 28 \\ 20 + s_1 = 28 \\ s_1 = 8 $$ 9. **Write the basic solution (I):** $$ (x_1, x_2, s_1, s_2, s_3) = (14, 6, 8, 0, 0) $$ **Final answer:** Basic solution (I) is $(14, 6, 8, 0, 0)$.