1. **State the problem:** Solve the rational equation $$x + \frac{3}{x} = 4$$ and determine which given options are solutions. 2. **Write the equation:** $$x + \frac{3}{x} = 4$$ 3. **Multiply both sides by $x$ to clear the denominator:** $$x \cdot x + x \cdot \frac{3}{x} = 4x$$ $$x^2 + \cancel{x} \cdot \frac{3}{\cancel{x}} = 4x$$ $$x^2 + 3 = 4x$$ 4. **Rearrange to form a quadratic equation:** $$x^2 - 4x + 3 = 0$$ 5. **Factor the quadratic:** $$x^2 - 4x + 3 = (x - 3)(x - 1) = 0$$ 6. **Solve for $x$:** $$x - 3 = 0 \Rightarrow x = 3$$ $$x - 1 = 0 \Rightarrow x = 1$$ 7. **Check for restrictions:** Since the original equation has $\frac{3}{x}$, $x \neq 0$ which is satisfied by both solutions. 8. **Conclusion:** Both $x=1$ and $x=3$ satisfy the equation. **Final answer:** Option A. $x=1$ or $x=3$.