1. **State the problem:** Find the values of $x$ on the real number line that satisfy the inequality $-x + 2 > 5$ and/or $5 \leq x + 4$. 2. **Solve the first inequality:** $$-x + 2 > 5$$ Subtract 2 from both sides: $$-x + \cancel{2} - \cancel{2} > 5 - 2$$ $$-x > 3$$ Multiply both sides by $-1$ and reverse the inequality sign (because multiplying by a negative reverses inequality): $$x < -3$$ 3. **Solve the second inequality:** $$5 \leq x + 4$$ Subtract 4 from both sides: $$5 - 4 \leq x + \cancel{4} - \cancel{4}$$ $$1 \leq x$$ or equivalently $$x \geq 1$$ 4. **Combine the inequalities with "and/or":** The solution is all $x$ such that $x < -3$ or $x \geq 1$. 5. **Interpretation:** - The first inequality corresponds to all real numbers less than $-3$. - The second inequality corresponds to all real numbers greater than or equal to $1$. 6. **Check the options:** - Option (A) shows $(1,4]$ which is part of the solution but misses $x < -3$. - Option (B) shows $(-1,3]$ which does not include $x < -3$ or $x \geq 1$ fully. - Option (C) shows $(-1,4]$ which misses $x < -3$. - Option (D) same as (B). - Option (E) shows $[-4,1)$ which includes $x < -3$ but excludes $x \geq 1$. Since the solution is $x < -3$ or $x \geq 1$, the correct intervals are $(-\infty, -3)$ and $[1, \infty)$. None of the options exactly match this, but option (E) covers $[-4,1)$ which includes $x < -3$ but excludes $x \geq 1$. Therefore, the correct solution set is $x < -3$ or $x \geq 1$. **Final answer:** $$\boxed{x < -3 \text{ or } x \geq 1}$$