If $g(2x - 5) = 3x + 9$ for all real $x$, then what is $g(1)$?
In order to use $g(2x-5) = 3x + 9$ to evaluate $g(1)$, we find the value of $x$ such that $2x-5 =1$.  Solving this equation gives $x=3$, so letting $x=3$ in $g(2x-5) = 3x+9$ gives $g(1) = \boxed{18}$.
18
Level 4
Algebra
Problem: If $g(2x - 5) = 3x + 9$ for all real $x$, then what is $g(1)$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \( g(1) \), we first need to determine the function \( g \) from the equation \( g(2x - 5) = 3x + 9 \). 

1. Set \( y = 2x - 5 \). This implies \( x = \frac{y + 5}{2} \).
2. Substitute \( x \) back into the equation: 
   \[
   g(y) = 3\left(\frac{y + 5}{2}\right) + 9
   \]
3. Simplify the right side:
   \[
   g(y) = \frac{3(y + 5)}{2} + 9 = \frac{3y + 15}{2} + 9
   \]
4. Convert \( 9 \) to a fraction with a denominator of 2:
   \[
   g(y) = \frac{3y + 15}{2} + \frac{18}{2} = \frac{3y + 33}{2}
   \]
5. Thus, we have \( g(y) = \frac{3y + 33}{2} \).

Next, we need to find \( g(1) \):
1. Substitute \( y = 1 \) into the function:
   \[
   g(1) = \frac{3(1) + 33}{2} = \frac{3 + 33}{2} = \frac{36}{2} = 18
   \]

Therefore, the final answer is \( 18 \).<|END_OF_LATENT|> The final answer is $18$.