1. **State the problem:** Solve the equation $$\frac{4r-3}{6r+1} = \frac{2r-1}{3r+4}$$ for $r$. 2. **Cross-multiply to eliminate the fractions:** $$ (4r-3)(3r+4) = (2r-1)(6r+1) $$ 3. **Expand both sides:** $$ (4r)(3r) + (4r)(4) - 3(3r) - 3(4) = (2r)(6r) + (2r)(1) - 1(6r) - 1(1) $$ $$ 12r^2 + 16r - 9r - 12 = 12r^2 + 2r - 6r - 1 $$ 4. **Simplify both sides:** $$ 12r^2 + 7r - 12 = 12r^2 - 4r - 1 $$ 5. **Subtract $12r^2$ from both sides:** $$ 7r - 12 = -4r - 1 $$ 6. **Add $4r$ to both sides:** $$ 7r + 4r - 12 = -1 $$ $$ 11r - 12 = -1 $$ 7. **Add 12 to both sides:** $$ 11r = 11 $$ 8. **Divide both sides by 11:** $$ r = 1 $$ **Final answer:** $$r = 1$$