1. **State the problem:** A bag contains 12 marbles: 5 orange, 3 blue, and the rest red. Two marbles are selected without replacement. Find the probability that one marble is red and the other is blue. 2. **Identify the number of red marbles:** Total marbles = 12, orange = 5, blue = 3, so red = 12 - 5 - 3 = 4. 3. **Total ways to choose 2 marbles:** $$\binom{12}{2} = \frac{12 \times 11}{2} = 66$$ 4. **Ways to choose 1 red and 1 blue marble:** Choose 1 red from 4 and 1 blue from 3: $$\binom{4}{1} \times \binom{3}{1} = 4 \times 3 = 12$$ 5. **Calculate the probability:** $$P = \frac{12}{66} = \frac{\cancel{12}}{\cancel{66}} = \frac{2}{11}$$ --- 6. **Solve for x:** Given equation: $$\frac{3}{5}x - \frac{4}{15} = \frac{1}{3}(2x - 5)$$ 7. **Distribute right side:** $$\frac{3}{5}x - \frac{4}{15} = \frac{2}{3}x - \frac{5}{3}$$ 8. **Bring all terms to one side:** $$\frac{3}{5}x - \frac{2}{3}x = -\frac{5}{3} + \frac{4}{15}$$ 9. **Find common denominators and simplify left side:** $$\frac{9}{15}x - \frac{10}{15}x = -\frac{5}{3} + \frac{4}{15}$$ $$\cancel{\frac{9}{15}}x - \cancel{\frac{10}{15}}x = -\frac{5}{3} + \frac{4}{15}$$ $$-\frac{1}{15}x = -\frac{5}{3} + \frac{4}{15}$$ 10. **Simplify right side:** Convert $-\frac{5}{3}$ to fifteenths: $$-\frac{5}{3} = -\frac{25}{15}$$ So, $$-\frac{25}{15} + \frac{4}{15} = -\frac{21}{15} = -\frac{7}{5}$$ 11. **Equation now:** $$-\frac{1}{15}x = -\frac{7}{5}$$ 12. **Multiply both sides by $-15$ to solve for x:** $$x = -\frac{7}{5} \times (-15) = \frac{7}{5} \times 15 = 7 \times 3 = 21$$ --- **Final answers:** - Probability that one marble is red and the other blue: $$\frac{2}{11}$$ - Value of $$x$$: $$21$$