1. Stating the problem: Solve the system of inequalities (a) $3x + 2x(x - 3) \leq 2(5 + x^2)$ (b) $6 - 2 + 3x + 4x \leq 9$ 2. Solve inequality (a): Start by expanding and simplifying both sides. Left side: $3x + 2x(x - 3) = 3x + 2x^2 - 6x = 2x^2 - 3x$ Right side: $2(5 + x^2) = 10 + 2x^2$ Rewrite inequality: $$2x^2 - 3x \leq 10 + 2x^2$$ Subtract $2x^2$ from both sides: $$2x^2 - 3x - 2x^2 \leq 10 + 2x^2 - 2x^2 \Rightarrow -3x \leq 10$$ Divide both sides by $-3$ and reverse inequality sign (dividing by negative): $$x \geq -\frac{10}{3}$$ 3. Solve inequality (b): Simplify left side: $$6 - 2 + 3x + 4x = 4 + 7x$$ Inequality becomes: $$4 + 7x \leq 9$$ Subtract 4 from both sides: $$7x \leq 5$$ Divide both sides by 7: $$x \leq \frac{5}{7}$$ 4. Combine solutions for the system: $$x \geq -\frac{10}{3} \quad \text{and} \quad x \leq \frac{5}{7}$$ Therefore, the solution set is: $$\boxed{-\frac{10}{3} \leq x \leq \frac{5}{7}}$$