In a bag, there are a total of 15 balls consisting of red and blue balls. If two balls are drawn randomly, the probability that both balls are of the same color is $\frac{17}{35}$. Given that the number of blue balls is greater than the number of red balls, find the number of blue balls.

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**Solution:**

1. Let the number of blue balls be $x$. Then the number of red balls is $15 - x$.

2. The total number of ways to pick any 2 balls from 15 is:
   $$\binom{15}{2} = \frac{15 	imes 14}{2} = 105$$

3. The number of ways to pick 2 blue balls is:
   $$\binom{x}{2} = \frac{x(x-1)}{2}$$

4. The number of ways to pick 2 red balls is:
   $$\binom{15 - x}{2} = \frac{(15 - x)(14 - x)}{2}$$

5. The probability of picking two balls of the same color is given as:
   $$\frac{\binom{x}{2} + \binom{15 - x}{2}}{\binom{15}{2}} = \frac{17}{35}$$

6. Substitute the values:
   $$\frac{\frac{x(x-1)}{2} + \frac{(15 - x)(14 - x)}{2}}{105} = \frac{17}{35}$$

7. Multiply both sides by 105:
   $$\frac{x(x-1)}{2} + \frac{(15 - x)(14 - x)}{2} = 105 	imes \frac{17}{35} = 3 	imes 17 = 51$$

8. Multiply both sides by 2:
   $$x(x-1) + (15 - x)(14 - x) = 102$$

9. Expand the terms:
   $$x^2 - x + (15 - x)(14 - x) = 102$$

Expand $(15 - x)(14 - x)$:
   $$= 15 	imes 14 - 15x - 14x + x^2 = 210 - 29x + x^2$$

10. Substitute back:
    $$x^2 - x + 210 - 29x + x^2 = 102$$

Combine like terms:
    $$2x^2 - 30x + 210 = 102$$

11. Simplify:
    $$2x^2 - 30x + 210 - 102 = 0$$
    $$2x^2 - 30x + 108 = 0$$

12. Divide entire equation by 2:
    $$x^2 - 15x + 54 = 0$$

13. Factorize the quadratic:
    $$(x - 9)(x - 6) = 0$$

14. So, $x = 9$ or $x = 6$.

15. Since the number of blue balls is greater than the number of red balls, and total balls are 15, blue balls must be $9$ (because if $x=6$, blue balls are fewer).

**Answer:** $9$

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