1. **Stating the problem:** Solve the system of inequalities: $$\begin{cases} x^2 + 3x - 4 < 0 \\ 1 - 3x > 0 \end{cases}$$ 2. **Solve the first inequality:** $$x^2 + 3x - 4 < 0$$ We start by finding the roots of the quadratic equation $x^2 + 3x - 4 = 0$ using the discriminant formula: $$\Delta = b^2 - 4ac = 3^2 - 4 \times 1 \times (-4) = 9 + 16 = 25$$ The roots are: $$x_1 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-3 - 5}{2} = \frac{-8}{2} = -4$$ $$x_2 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-3 + 5}{2} = \frac{2}{2} = 1$$ 3. **Determine the sign of the quadratic:** Since the parabola opens upwards ($a=1 > 0$), the quadratic is less than zero between the roots: $$-4 < x < 1$$ 4. **Solve the second inequality:** $$1 - 3x > 0$$ Rearranging: $$-3x > -1$$ Dividing both sides by $-3$ (remember to reverse the inequality sign when dividing by a negative): $$\cancel{-3}x < \cancel{-1} \Rightarrow x < \frac{1}{3}$$ 5. **Combine the two inequalities:** We want values of $x$ that satisfy both: $$-4 < x < 1$$ and $$x < \frac{1}{3}$$ The intersection is: $$-4 < x < \frac{1}{3}$$ **Final answer:** $$\boxed{-4 < x < \frac{1}{3}}$$