1. **Problem 19:** Solve the equation $$\sqrt{y - 5} + 5 = y$$ 2. **Step 1:** Isolate the square root term: $$\sqrt{y - 5} = y - 5$$ 3. **Step 2:** Square both sides to eliminate the square root: $$\left(\sqrt{y - 5}\right)^2 = (y - 5)^2$$ $$y - 5 = (y - 5)^2$$ 4. **Step 3:** Expand the right side: $$(y - 5)^2 = y^2 - 10y + 25$$ 5. **Step 4:** Set up the quadratic equation: $$y - 5 = y^2 - 10y + 25$$ 6. **Step 5:** Bring all terms to one side: $$0 = y^2 - 10y + 25 - y + 5$$ $$0 = y^2 - 11y + 30$$ 7. **Step 6:** Factor the quadratic: $$(y - 5)(y - 6) = 0$$ 8. **Step 7:** Solve for $y$: $$y = 5 \quad \text{or} \quad y = 6$$ 9. **Step 8:** Check for extraneous solutions by substituting back into the original equation: - For $y=5$: $$\sqrt{5 - 5} + 5 = 0 + 5 = 5$$ Right side: $y=5$; valid. - For $y=6$: $$\sqrt{6 - 5} + 5 = \sqrt{1} + 5 = 1 + 5 = 6$$ Right side: $y=6$; valid. **Both solutions are valid.** --- 10. **Problem 21:** Solve the equation $$\sqrt{8x + 1} + 4 = 0$$ 11. **Step 1:** Isolate the square root term: $$\sqrt{8x + 1} = -4$$ 12. **Step 2:** Note that the square root function always returns a non-negative value, so $$\sqrt{8x + 1} \geq 0$$. 13. **Step 3:** Since the right side is negative, there is **no solution** to this equation. **Final answers:** - For problem 19: $$y = 5 \text{ or } y = 6$$ - For problem 21: No solution.