1. **Problem statement:** Find the domain of the function $$y = \sqrt{4x - 3} + \frac{1}{6 - x}.$$\n\n2. **Recall domain rules:**\n- The expression under the square root must be non-negative: $$4x - 3 \geq 0.$$\n- The denominator cannot be zero: $$6 - x \neq 0.$$\n\n3. **Solve the inequality for the square root:**\n$$4x - 3 \geq 0 \implies 4x \geq 3 \implies x \geq \frac{3}{4}.$$\n\n4. **Solve the denominator restriction:**\n$$6 - x \neq 0 \implies x \neq 6.$$\n\n5. **Combine the conditions:**\nThe domain is all $$x$$ such that $$x \geq \frac{3}{4}$$ and $$x \neq 6.$$\n\n6. **Write the domain in interval notation:**\n$$\left[\frac{3}{4}, 6\right) \cup (6, +\infty).$$