1. **State the problem:** Simplify and solve the equation $$\frac{3x - 1}{6x - 3} + \frac{1}{4x^2 - 1} = \frac{-x}{2x + 1}.$$\n\n2. **Identify and factor denominators:**\n- $6x - 3 = 3(2x - 1)$\n- $4x^2 - 1 = (2x - 1)(2x + 1)$ (difference of squares)\n- $2x + 1$ is already factored\n\n3. **Rewrite the equation with factored denominators:**\n$$\frac{3x - 1}{3(2x - 1)} + \frac{1}{(2x - 1)(2x + 1)} = \frac{-x}{2x + 1}.$$\n\n4. **Find the least common denominator (LCD):**\nThe LCD is $3(2x - 1)(2x + 1)$.\n\n5. **Multiply both sides of the equation by the LCD to clear denominators:**\n$$\left(\frac{3x - 1}{3(2x - 1)}\right) \times 3(2x - 1)(2x + 1) + \left(\frac{1}{(2x - 1)(2x + 1)}\right) \times 3(2x - 1)(2x + 1) = \left(\frac{-x}{2x + 1}\right) \times 3(2x - 1)(2x + 1).$$\n\n6. **Simplify each term:**\n- First term: $(3x - 1)(2x + 1)$\n- Second term: $3$\n- Right side: $-x \times 3(2x - 1) = -3x(2x - 1)$\n\n7. **Write the simplified equation:**\n$$(3x - 1)(2x + 1) + 3 = -3x(2x - 1).$$\n\n8. **Expand the products:**\n- Left: $(3x)(2x) + (3x)(1) - 1(2x) - 1(1) + 3 = 6x^2 + 3x - 2x - 1 + 3 = 6x^2 + x + 2$\n- Right: $-3x(2x) + 3x = -6x^2 + 3x$\n\n9. **Set up the equation:**\n$$6x^2 + x + 2 = -6x^2 + 3x.$$\n\n10. **Bring all terms to one side:**\n$$6x^2 + x + 2 + 6x^2 - 3x = 0 \implies 12x^2 - 2x + 2 = 0.$$\n\n11. **Simplify by dividing all terms by 2:**\n$$6x^2 - x + 1 = 0.$$\n\n12. **Use the quadratic formula:**\n$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$ where $a=6$, $b=-1$, $c=1$.\n\n13. **Calculate the discriminant:**\n$$\Delta = (-1)^2 - 4 \times 6 \times 1 = 1 - 24 = -23.$$\n\n14. **Since the discriminant is negative, there are no real solutions.**\n\n**Final answer:** The equation has no real solutions because the discriminant is negative.