1. **Stating the problem:** Solve the system of inequalities: $$x^2 + 3x - 4 < 0$$ $$1 - 3x \geq 0$$ 2. **Solve the quadratic inequality $x^2 + 3x - 4 < 0$: ** First, find 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 intervals where $x^2 + 3x - 4 < 0$: ** Since the parabola opens upwards ($a=1 > 0$), the quadratic is less than zero between the roots: $$-4 < x < 1$$ 4. **Solve the linear inequality $1 - 3x \geq 0$: ** Rearrange: $$1 \geq 3x$$ Divide both sides by 3 (positive, so inequality direction stays the same): $$\frac{1}{3} \geq x$$ Or equivalently: $$x \leq \frac{1}{3}$$ 5. **Combine both inequalities: ** We want values of $x$ satisfying both: $$-4 < x < 1$$ and $$x \leq \frac{1}{3}$$ The intersection is: $$-4 < x \leq \frac{1}{3}$$ **Final answer:** $$\boxed{-4 < x \leq \frac{1}{3}}$$