1. **State the problem:** We need to find the solutions to the system of equations represented by the two curves: a concave down parabola and a straight line with negative slope. 2. **Identify the equations:** From the graph description, the parabola peaks around $x=4$ and extends roughly from $x=1$ to $x=7$. The line crosses the y-axis near 10 and descends with negative slope. 3. **Set up the system:** Let the parabola be $y = -a(x - 4)^2 + b$ for some positive $a$ and $b$ (since it is concave down and peaks at $x=4$). The line can be written as $y = mx + c$ with $m < 0$ and $c \approx 10$. 4. **Find intersection points:** The solutions to the system are the $x$ values where the parabola and line intersect, i.e., where $$-a(x - 4)^2 + b = mx + c.$$ 5. **Rewrite the equation:** Move all terms to one side: $$-a(x - 4)^2 + b - mx - c = 0.$$ 6. **Expand and simplify:** $$-a(x^2 - 8x + 16) + b - mx - c = 0$$ $$-a x^2 + 8a x - 16a + b - mx - c = 0$$ 7. **Group like terms:** $$-a x^2 + (8a - m) x + (b - 16a - c) = 0.$$ 8. **Solve quadratic:** The number of solutions depends on the discriminant $$\Delta = (8a - m)^2 - 4(-a)(b - 16a - c).$$ 9. **Interpretation:** If $\Delta > 0$, two solutions; if $\Delta = 0$, one solution; if $\Delta < 0$, no real solutions. 10. **From the graph:** The parabola and line intersect twice, so the system has two solutions. **Final answer:** The system of equations has two solutions corresponding to the two intersection points of the parabola and the line on the graph.