Suppose $\sqrt{1 + \sqrt{2y-3}} = \sqrt{6}$; find $y$.
This is an equation involving nested square roots, so I want to simplify it by squaring both sides and isolating the inner square root. Squaring both sides gives me $1 + \sqrt{2y-3} = 6$. Subtracting 1 from both sides gives me $\sqrt{2y-3} = 5$. Now I have a single square root to deal with, so I square both sides again to get rid of it. Squaring both sides gives me $2y-3 = 25$. Adding 3 to both sides gives me $2y = 28$. Dividing both sides by 2 gives me $y = 14$. To check my answer, I can plug it back into the original equation and see if it makes sense. If $y = 14$, then $\sqrt{1 + \sqrt{2y-3}} = \sqrt{1 + \sqrt{2(14)-3}} = \sqrt{1 + \sqrt{25}} = \sqrt{1 + 5} = \sqrt{6}$, which is correct. So $y = 14$ is the solution. 