Question: User: $\frac{6r - 5}{9r - 3} = - \frac{1}{3}$
1. **State the problem:** Solve the equation $$\frac{6r - 5}{9r - 3} = - \frac{1}{3}$$ for $r$.
2. **Use the cross-multiplication method:** When two fractions are equal, their cross products are equal. So,
$$ (6r - 5) \times 3 = (9r - 3) \times (-1) $$
3. **Multiply out both sides:**
$$ 3(6r - 5) = -1(9r - 3) $$
$$ 18r - 15 = -9r + 3 $$
4. **Bring all terms involving $r$ to one side and constants to the other:**
$$ 18r + 9r = 3 + 15 $$
$$ 27r = 18 $$
5. **Solve for $r$ by dividing both sides by 27:**
$$ r = \frac{18}{27} $$
6. **Simplify the fraction by canceling common factors:**
$$ r = \frac{\cancel{18}^{2 \times 9}}{\cancel{27}^{3 \times 9}} = \frac{2}{3} $$
**Final answer:** $$r = \frac{2}{3}$$