1. The problem asks to find the real solutions of the inequality $x - 3 \leq 2$ where $x$ is the real part of the complex number $z = x + ij$. 2. To solve the inequality, isolate $x$ by adding 3 to both sides: $$x - 3 \leq 2 \implies x \leq 2 + 3$$ 3. Simplifying the right side: $$x \leq 5$$ 4. Therefore, the real solutions are all real numbers $x$ such that $x \leq 5$. Final answer: All real $x$ where $x \leq 5$.