1. **State the problem:** We have a table with variables $x$, $y$, $u_1$, $u_2$, and a value $z$ to find in the last row. The table is: | $x$ | $y$ | $u_1$ | $u_2$ | Value | |-----|-----|-------|-------|-------| | 2 | 2 | 1 | 0 | 6 | | 2 | 1 | 0 | 1 | 5 | | 6 | 4 | 0 | 0 | $z$ | We want to find $z$. 2. **Analyze the problem:** The columns $u_1$ and $u_2$ seem to be indicator variables or weights for the values 6 and 5 in the first two rows. The last row has zeros for $u_1$ and $u_2$, so $z$ might be related to $x$ and $y$ directly. 3. **Check if $z$ can be expressed as a linear combination:** Since $u_1$ and $u_2$ are zero in the last row, $z$ is not directly from those. Let's check if $z$ relates to $x$ and $y$ by a linear formula: Assume $z = a x + b y$. 4. **Use the first two rows to find $a$ and $b$:** From row 2: $6 = a \cdot 2 + b \cdot 2$ From row 3: $5 = a \cdot 2 + b \cdot 1$ 5. **Solve the system:** Subtract the second equation from the first: $$6 - 5 = (a \cdot 2 + b \cdot 2) - (a \cdot 2 + b \cdot 1)$$ $$1 = b (2 - 1)$$ $$1 = b$$ 6. **Find $a$:** Use $b=1$ in second equation: $$5 = 2a + 1 \cdot 1$$ $$5 = 2a + 1$$ $$5 - 1 = 2a$$ $$4 = 2a$$ $$a = 2$$ 7. **Calculate $z$ for the last row:** $$z = 2 \cdot 6 + 1 \cdot 4 = 12 + 4 = 16$$ **Final answer:** $z = 16$