1. **State the problem:** Simplify the expression $$\frac{6x - 1}{3x - 4} + \frac{4x - 1}{4 - 3x}$$ given the restrictions $$x \neq 4$$ and $$x \neq -1$$. 2. **Identify the restrictions:** The denominators cannot be zero, so: - $$3x - 4 \neq 0 \Rightarrow x \neq \frac{4}{3}$$ - $$4 - 3x \neq 0 \Rightarrow x \neq \frac{4}{3}$$ (same restriction) - Also given $$x \neq 4$$ and $$x \neq -1$$. 3. **Rewrite the second denominator:** Note that $$4 - 3x = -(3x - 4)$$. 4. **Rewrite the second fraction:** $$\frac{4x - 1}{4 - 3x} = \frac{4x - 1}{-(3x - 4)} = -\frac{4x - 1}{3x - 4}$$ 5. **Combine the fractions:** $$\frac{6x - 1}{3x - 4} + \left(-\frac{4x - 1}{3x - 4}\right) = \frac{6x - 1 - (4x - 1)}{3x - 4}$$ 6. **Simplify the numerator:** $$6x - 1 - 4x + 1 = (6x - 4x) + (-1 + 1) = 2x + 0 = 2x$$ 7. **Simplify the expression:** $$\frac{2x}{3x - 4}$$ 8. **State the final restrictions:** - $$x \neq \frac{4}{3}$$ (denominator zero) - $$x \neq 4$$ and $$x \neq -1$$ (given) **Final answer:** $$\boxed{\frac{2x}{3x - 4}}$$ with restrictions $$x \neq \frac{4}{3}, x \neq 4, x \neq -1$$.