1. **State the problem:** Solve the equation $$ff(x) = gf(2)$$ where $$f(x) = 10 - 3x$$ and $$g(x) = \frac{10}{3 - 2x}$$ with domain restrictions $$x \in \mathbb{R}$$ and $$x \neq \frac{3}{2}$$. 2. **Find $$ff(x)$$:** This means applying $$f$$ twice: $$ff(x) = f(f(x))$$. Calculate $$f(x)$$ first: $$f(x) = 10 - 3x$$ Now apply $$f$$ again to $$f(x)$$: $$ff(x) = f(10 - 3x) = 10 - 3(10 - 3x)$$ Simplify inside the parentheses: $$ff(x) = 10 - 3 \times 10 + 3 \times 3x = 10 - 30 + 9x$$ Simplify further: $$ff(x) = -20 + 9x$$ 3. **Calculate $$gf(2)$$:** First find $$f(2)$$: $$f(2) = 10 - 3 \times 2 = 10 - 6 = 4$$ Now find $$g(f(2)) = g(4)$$: $$g(4) = \frac{10}{3 - 2 \times 4} = \frac{10}{3 - 8} = \frac{10}{-5} = -2$$ 4. **Set up the equation:** $$ff(x) = gf(2)$$ $$-20 + 9x = -2$$ 5. **Solve for $$x$$:** Add 20 to both sides: $$-20 + 9x + 20 = -2 + 20$$ $$9x = 18$$ Divide both sides by 9: $$\frac{\cancel{9}x}{\cancel{9}} = \frac{18}{9}$$ $$x = 2$$ 6. **Check domain restrictions:** $$x = 2$$ is in $$\mathbb{R}$$ and $$2 \neq \frac{3}{2}$$, so it is valid. **Final answer:** $$\boxed{2}$$