Suppose $\sqrt{1 + \sqrt{2y-3}} = \sqrt{6}$; find $y$.
Squaring both sides, $$1 + \sqrt{2y-3} = \left(\sqrt{1 + \sqrt{2y-3}}\right)^2 = \left(\sqrt{6}\right)^2 = 6.$$Hence, $\sqrt{2y-3} = 5$. If we square this equation again, then $$2y - 3 = \left(\sqrt{2y-3}\right)^2 = 5^2 = 25 \Longrightarrow y = \frac{25+3}{2} = \boxed{14}.$$
14
Level 3
Algebra
Problem: Suppose $\sqrt{1 + \sqrt{2y-3}} = \sqrt{6}$; find $y$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation $\sqrt{1 + \sqrt{2y-3}} = \sqrt{6}$, we first square both sides to eliminate the outer square root. This gives us:

1 + \sqrt{2y-3} = 6.

Next, we isolate the inner square root by subtracting 1 from both sides:

\sqrt{2y-3} = 5.

We then square both sides again to eliminate the square root:

2y - 3 = 25.

Now, we solve for y by first adding 3 to both sides:

2y = 28.

Finally, we divide both sides by 2 to find y:

y = 14.

Thus, the final answer is 14.<|END_OF_LATENT|> The final answer is $14$.