1. **State the problem:** We need to find the three inequalities that define the unshaded region on the graph. 2. **Identify the lines and their equations:** - Solid sloped line with endpoints (0,16) and (6,0) has slope $m = \frac{0-16}{6-0} = -\frac{16}{6} = -\frac{8}{3}$, but the problem states approximately $y = -2x + 16$ which is close and simpler. - Dashed line passing through (0,3) and (6,9) has slope $m = \frac{9-3}{6-0} = 1$, so equation is $y = x + 3$. - Vertical solid line at $x = 4$. 3. **Determine inequalities for the unshaded region:** - The unshaded region is to the right of the vertical line $x=4$, so $x \geq 4$. - It is below the dashed line $y = x + 3$, so $y \leq x + 3$. - It is above the solid sloped line $y = -2x + 16$, so $y \geq -2x + 16$. 4. **Write the inequalities:** $$ \begin{cases} x \geq 4 \\ y \leq x + 3 \\ y \geq -2x + 16 \end{cases} $$ These three inequalities define the unshaded region. **Final answer:** $$x \geq 4, \quad y \leq x + 3, \quad y \geq -2x + 16$$