1. **State the problem:** Determine if the functions $f(x) = (x-1)(x-2)$ and $g(x) = \frac{x-1}{x-2}$ are linearly independent. 2. **Recall the definition:** Two functions $f$ and $g$ are linearly independent if the only solution to $c_1 f(x) + c_2 g(x) = 0$ for all $x$ is $c_1 = 0$ and $c_2 = 0$. 3. **Set up the equation:** $$c_1 (x-1)(x-2) + c_2 \frac{x-1}{x-2} = 0$$ for all $x$ where $x \neq 2$ (to avoid division by zero). 4. **Multiply both sides by $(x-2)$ to clear the denominator:** $$c_1 (x-1)(x-2) \cancel{(x-2)} + c_2 \frac{x-1}{x-2} \cancel{(x-2)} = 0 \cdot (x-2)$$ which simplifies to $$c_1 (x-1)(x-2)^2 + c_2 (x-1) = 0$$ 5. **Factor out $(x-1)$:** $$ (x-1) \left[c_1 (x-2)^2 + c_2 \right] = 0$$ 6. **For this to hold for all $x$, the expression inside brackets must be zero for all $x \neq 1$:** $$c_1 (x-2)^2 + c_2 = 0$$ 7. **Since $(x-2)^2$ is a quadratic function, the only way this holds for all $x$ is if both coefficients are zero:** $$c_1 = 0 \quad \text{and} \quad c_2 = 0$$ 8. **Conclusion:** The only solution is the trivial one, so $f$ and $g$ are linearly independent. **Final answer:** The functions $(x-1)(x-2)$ and $\frac{x-1}{x-2}$ are linearly independent.