Question: The partially completed table below shows the ten basic solutions to the following e-system. $x_1 + x_2 + s_1 = 29$ $2x_1 + x_2 + s_2 = 30$ $4x_1 + x_2 + s_3 = 44$ Find basic solution (I). (A) 0 0 29 30 44 (B) 0 29 0 1 15 (C) 0 30 -1 0 14 (D) 0 44 -15 -14 0 (E) 29 0 0 -28 -72 (F) 15 0 14 0 -16 (G) (H) (I) (J) What is basic solution (I)? $(x_1, x_2, s_1, s_2, s_3) = (__, __, 0, __, 0)

1. **State the problem:** We are given the system of equations: $$x_1 + x_2 + s_1 = 29$$ $$2x_1 + x_2 + s_2 = 30$$ $$4x_1 + x_2 + s_3 = 44$$ and we want to find the basic solution (I) where $s_1 = 0$ and $s_3 = 0$. 2. **Set the known values:** Since $s_1 = 0$ and $s_3 = 0$, substitute these into the system: $$x_1 + x_2 = 29$$ $$2x_1 + x_2 + s_2 = 30$$ $$4x_1 + x_2 = 44$$ 3. **Express $s_2$ from the second equation:** $$s_2 = 30 - 2x_1 - x_2$$ 4. **Use the first and third equations to solve for $x_1$ and $x_2$:** From the first: $$x_2 = 29 - x_1$$ Substitute into the third: $$4x_1 + (29 - x_1) = 44$$ Simplify: $$4x_1 + 29 - x_1 = 44$$ $$3x_1 + 29 = 44$$ $$3x_1 = 44 - 29$$ $$3x_1 = 15$$ $$x_1 = \frac{15}{3} = 5$$ 5. **Find $x_2$:** $$x_2 = 29 - 5 = 24$$ 6. **Find $s_2$:** $$s_2 = 30 - 2(5) - 24 = 30 - 10 - 24 = -4$$ 7. **Write the basic solution (I):** $$(x_1, x_2, s_1, s_2, s_3) = (5, 24, 0, -4, 0)$$ Note: $s_2$ is negative, which may indicate this is a non-feasible basic solution, but it satisfies the system with the given zero slack variables. **Final answer:** $(5, 24, 0, -4, 0)$