1. **State the problem:** We have two sets defined by equations: Set A: $y=\frac{2x-1}{x}$ Set B: $y=ax+2$ We are told that the intersection $A \cap B$ is non-defined, meaning there is no solution $(x,y)$ that satisfies both equations simultaneously. 2. **Set the equations equal to find intersection:** To find the intersection, set the expressions for $y$ equal: $$\frac{2x-1}{x} = ax + 2$$ 3. **Multiply both sides by $x$ to clear the denominator:** $$2x - 1 = ax^2 + 2x$$ 4. **Rearrange to form a quadratic equation:** $$ax^2 + 2x - (2x - 1) = 0$$ Simplify the right side: $$ax^2 + 2x - 2x + 1 = 0$$ $$ax^2 + 1 = 0$$ 5. **Rewrite the quadratic:** $$ax^2 = -1$$ 6. **Analyze the quadratic for real solutions:** For $x$ to be real, the equation $ax^2 = -1$ must have real solutions. Since $x^2 \geq 0$ for all real $x$, the right side $-1$ is negative. Therefore, for any real $x$, $ax^2 = -1$ implies $a$ must be negative to make the left side negative. 7. **Check if solutions exist:** Rewrite as: $$x^2 = -\frac{1}{a}$$ For $x^2$ to be real and non-negative, $-\frac{1}{a} \geq 0$. This means: $$-\frac{1}{a} \geq 0 \implies a < 0$$ 8. **Condition for no intersection:** If $a < 0$, then $x^2 = -\frac{1}{a} > 0$ and there are two real solutions, so intersection exists. If $a > 0$, then $x^2 = -\frac{1}{a} < 0$ which is impossible for real $x$, so no real solutions exist. 9. **Conclusion:** The intersection is non-defined (no real solutions) if and only if: $$a > 0$$ **Final answer:** $$\boxed{a > 0}$$