1. **State the problem:** Solve the inequality $$-3x + 10 \leq 19$$ for $$x$$. 2. **Isolate the variable term:** Subtract 10 from both sides: $$-3x + 10 - 10 \leq 19 - 10$$ $$-3x \leq 9$$ 3. **Divide both sides by -3:** Since we are dividing by a negative number, the inequality sign reverses: $$\frac{\cancel{-3}x}{\cancel{-3}} \geq \frac{9}{-3}$$ $$x \geq -3$$ 4. **Final answer:** $$x \geq -3$$ This means $$x$$ can be any number greater than or equal to $$-3$$.