1. **State the problem:** Solve the system of inequalities: $$\frac{x-2}{x-4} < 0$$ and $$x^2 + 3x - 4 > 0$$ 2. **Solve the first inequality:** $$\frac{x-2}{x-4} < 0$$ - The expression changes sign at points where numerator or denominator is zero: at $x=2$ and $x=4$. - The denominator cannot be zero, so $x \neq 4$. - Test intervals: - For $x < 2$, numerator $(x-2)<0$, denominator $(x-4)<0$, so fraction is positive. - For $2 < x < 4$, numerator $(x-2)>0$, denominator $(x-4)<0$, fraction is negative. - For $x > 4$, numerator $(x-2)>0$, denominator $(x-4)>0$, fraction is positive. Thus, solution for first inequality is: $$2 < x < 4$$ 3. **Solve the second inequality:** $$x^2 + 3x - 4 > 0$$ - Factor the quadratic: $$x^2 + 3x - 4 = (x+4)(x-1)$$ - The quadratic is zero at $x=-4$ and $x=1$. - Since leading coefficient is positive, the parabola opens upward. - The quadratic is positive outside the roots: $$x < -4 \quad \text{or} \quad x > 1$$ 4. **Combine the solutions:** - From first inequality: $2 < x < 4$ - From second inequality: $x < -4$ or $x > 1$ - Intersection is where both are true: $$2 < x < 4 \quad \text{and} \quad x > 1 \implies 2 < x < 4$$ 5. **Final solution:** $$\boxed{2 < x < 4}$$ This is the set of $x$ values satisfying both inequalities simultaneously.