1. **Stating the problem:** Solve the given algebraic and coordinate geometry problems involving quadratic and quartic equations, inequalities, and systems of linear equations. 2. **Quadratic equation and discriminant:** For equations like $3x^2 + 2x - 1 = 0$, use the quadratic formula: $$x = \frac{-b \pm \sqrt{\Delta}}{2a}$$ where $\Delta = b^2 - 4ac$. 3. **Solving $3x^2 + 2x - 1 = 0$:** Calculate discriminant: $$\Delta = 2^2 - 4 \times 3 \times (-1) = 4 + 12 = 16$$ Roots: $$x = \frac{-2 \pm \sqrt{16}}{2 \times 3} = \frac{-2 \pm 4}{6}$$ So, $$x_1 = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}, \quad x_2 = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$$ 4. **Inequality $3x^2 + 2x - 1 \leq 0$:** Since $a=3 > 0$, parabola opens upward. The solution is between roots: $$-1 \leq x \leq \frac{1}{3}$$ 5. **Quartic equation substitution:** Given $3t^4 + 2t^2 - 1 = 0$, let $t = x^2$. Rewrite as quadratic in $t^2$: $$3t^2 + 2t - 1 = 0$$ Solve for $t$: $$\Delta = 2^2 - 4 \times 3 \times (-1) = 16$$ $$t = \frac{-2 \pm 4}{6}$$ So, $$t_1 = \frac{2}{6} = \frac{1}{3}, \quad t_2 = \frac{-6}{6} = -1$$ Since $t = x^2$, discard $t_2 = -1$ (no real solution). 6. **Solve $x^2 = \frac{1}{3}$:** $$x = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$$ 7. **System of linear equations:** $$\begin{cases} 3x + y = 1 \\ x + 2y = -3 \end{cases}$$ Multiply second equation by 3: $$3x + 6y = -9$$ Subtract first equation: $$3x + 6y - (3x + y) = -9 - 1$$ $$5y = -10 \Rightarrow y = -2$$ Substitute $y$ into first equation: $$3x - 2 = 1 \Rightarrow 3x = 3 \Rightarrow x = 1$$ 8. **Parametric form of line:** Given: $$x = -5 - 2t, \quad y = 1 + t, \quad t \in \mathbb{R}$$ This represents a line in parametric form. 9. **Check if parametric line satisfies $3x + y - 1 = 0$:** Substitute: $$3(-5 - 2t) + (1 + t) - 1 = -15 - 6t + 1 + t - 1 = -15 - 5t$$ For this to be zero: $$-15 - 5t = 0 \Rightarrow t = -3$$ At $t = -3$, point lies on the line. 10. **Summary of solutions:** - Quadratic roots: $x = \frac{1}{3}, -1$ - Inequality solution: $-1 \leq x \leq \frac{1}{3}$ - Quartic roots: $x = \pm \frac{\sqrt{3}}{3}$ - System solution: $(x,y) = (1, -2)$ - Parametric line passes through $(x,y) = (-5 - 2t, 1 + t)$ Final answers: $$\boxed{\begin{cases} x = \frac{1}{3}, -1 \\ -1 \leq x \leq \frac{1}{3} \\ x = \pm \frac{\sqrt{3}}{3} \\ (x,y) = (1, -2) \end{cases}}$$