Question: 13 A design uses a point $(x, y)$ satisfying all three conditions: $x \geq 1$, $y \geq 2$, $x + y \leq 7$. The boundary lines are drawn on the grid. Boundaries: $x = 1$; $y = 2$; $x + y = 7$. (a) Shade the unwanted regions. Label the region that satisfies all three conditions $R$. (b) Can the point $(2, 3)$ be used? Give a reason.

1. **State the problem:** We need to find the region $R$ where all three inequalities $x \geq 1$, $y \geq 2$, and $x + y \leq 7$ hold simultaneously. 2. **Understand the boundaries:** - The line $x = 1$ is a vertical boundary; points to the right satisfy $x \geq 1$. - The line $y = 2$ is a horizontal boundary; points above satisfy $y \geq 2$. - The line $x + y = 7$ is a diagonal boundary; points below or on this line satisfy $x + y \leq 7$. 3. **Shade unwanted regions:** - Shade the region left of $x=1$ (where $x < 1$). - Shade the region below $y=2$ (where $y < 2$). - Shade the region above $x + y = 7$ (where $x + y > 7$). 4. **Label the feasible region $R$:** The intersection of the three conditions is the triangular region bounded by $x=1$, $y=2$, and $x + y=7$. 5. **Check if point $(2,3)$ can be used:** - Check $x \geq 1$: $2 \geq 1$ is true. - Check $y \geq 2$: $3 \geq 2$ is true. - Check $x + y \leq 7$: $2 + 3 = 5 \leq 7$ is true. Since all conditions are satisfied, the point $(2,3)$ lies in region $R$ and can be used. **Final answer:** - (a) The unwanted regions are left of $x=1$, below $y=2$, and above $x + y=7$. The feasible region $R$ is the triangle bounded by these lines. - (b) Yes, the point $(2,3)$ can be used because it satisfies all three inequalities.