1. **State the problem:** Solve the equation $$\sqrt{2x - 1} + \sqrt{x - 1} = 5$$ and verify the solutions. 2. **Rewrite the equation:** Let $$a = \sqrt{2x - 1}$$ and $$b = \sqrt{x - 1}$$, so the equation becomes $$a + b = 5$$. 3. **Express one variable in terms of the other:** $$a = 5 - b$$. 4. **Square both sides:** $$a^2 = (5 - b)^2$$ $$2x - 1 = 25 - 10b + b^2$$ 5. **Recall that $$b^2 = x - 1$$, substitute:** $$2x - 1 = 25 - 10b + (x - 1)$$ 6. **Simplify:** $$2x - 1 = 25 - 10b + x - 1$$ $$2x - 1 = x + 24 - 10b$$ 7. **Bring terms to one side:** $$2x - 1 - x - 24 = -10b$$ $$x - 25 = -10b$$ 8. **Divide both sides by -10:** $$\frac{x - 25}{-10} = b$$ Intermediate step with cancellation: $$b = \frac{\cancel{x - 25}}{\cancel{-10}}$$ (no common factors to cancel, so remains as is) 9. **Recall $$b = \sqrt{x - 1}$$, so:** $$\sqrt{x - 1} = \frac{25 - x}{10}$$ 10. **Square both sides again:** $$x - 1 = \left(\frac{25 - x}{10}\right)^2 = \frac{(25 - x)^2}{100}$$ 11. **Multiply both sides by 100:** $$100(x - 1) = (25 - x)^2$$ $$100x - 100 = (25 - x)^2$$ 12. **Expand the right side:** $$(25 - x)^2 = 625 - 50x + x^2$$ 13. **Set up the quadratic equation:** $$100x - 100 = 625 - 50x + x^2$$ 14. **Bring all terms to one side:** $$0 = 625 - 50x + x^2 - 100x + 100$$ $$0 = x^2 - 150x + 725$$ 15. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{150 \pm \sqrt{(-150)^2 - 4 \cdot 1 \cdot 725}}{2} = \frac{150 \pm \sqrt{22500 - 2900}}{2} = \frac{150 \pm \sqrt{19600}}{2}$$ 16. **Calculate the square root:** $$\sqrt{19600} = 140$$ 17. **Find the two solutions:** $$x_1 = \frac{150 + 140}{2} = \frac{290}{2} = 145$$ $$x_2 = \frac{150 - 140}{2} = \frac{10}{2} = 5$$ 18. **Verify solutions in the original equation:** - For $$x = 145$$: $$\sqrt{2(145) - 1} + \sqrt{145 - 1} = \sqrt{289} + \sqrt{144} = 17 + 12 = 29 \neq 5$$ (reject) - For $$x = 5$$: $$\sqrt{2(5) - 1} + \sqrt{5 - 1} = \sqrt{9} + \sqrt{4} = 3 + 2 = 5$$ (valid) **Final answer:** $$x = 5$$