1. **State the problem:** We need to solve the system of inequalities: $$x^2 + y - 8 \geq 0$$ $$y + 5x - 2 \geq 0$$ 2. **Rewrite each inequality to isolate $y$:** - For the quadratic inequality: $$x^2 + y - 8 \geq 0 \implies y \geq 8 - x^2$$ - For the linear inequality: $$y + 5x - 2 \geq 0 \implies y \geq 2 - 5x$$ 3. **Interpretation:** The solution set consists of all points $(x,y)$ where $y$ is greater than or equal to both $8 - x^2$ and $2 - 5x$ simultaneously. 4. **Find the intersection points of the boundary curves:** Set the right sides equal to find where the two curves meet: $$8 - x^2 = 2 - 5x$$ Rearranged: $$-x^2 + 5x + 6 = 0$$ Multiply both sides by $-1$: $$x^2 - 5x - 6 = 0$$ 5. **Solve the quadratic equation:** Using the quadratic formula: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm \sqrt{49}}{2}$$ So, $$x = \frac{5 + 7}{2} = 6$$ or $$x = \frac{5 - 7}{2} = -1$$ 6. **Find corresponding $y$ values:** For $x=6$: $$y = 8 - 6^2 = 8 - 36 = -28$$ For $x=-1$: $$y = 8 - (-1)^2 = 8 - 1 = 7$$ 7. **Solution region:** The solution is the set of points where: $$y \geq 8 - x^2$$ and $$y \geq 2 - 5x$$ which is the region above both curves. **Final answer:** $$\boxed{\{(x,y) \mid y \geq 8 - x^2 \text{ and } y \geq 2 - 5x\}}$$