Solve for $c$: \[\frac{c-23}{2} = \frac{2c +5}{7}.\]
Multiplying both sides by 2 and by 7 to eliminate fractions gives  \[7(c-23) = 2(2c+5).\] Expanding both sides gives $7c - 161 = 4c + 10$.  Subtracting $4c$ from both sides gives $3c -161= 10$, and adding 161 to both sides gives $3c = 171$.  Finally, dividing by 3 gives $c = \frac{171}{3} = \boxed{57}$.
57
Level 4
Prealgebra
Problem: Solve for $c$: \[\frac{c-23}{2} = \frac{2c +5}{7}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, first eliminate the fractions by finding a common denominator. The denominators are 2 and 7, so the least common multiple is 14. Multiply both sides of the equation by 14 to clear the fractions:

14 * \(\frac{c-23}{2}\) = 14 * \(\frac{2c + 5}{7}\).

This simplifies to:

7(c - 23) = 2(2c + 5).

Next, distribute on both sides:

7c - 161 = 4c + 10.

Now, isolate the variable \(c\) by moving all terms involving \(c\) to one side and constant terms to the other side. Subtract \(4c\) from both sides:

7c - 4c - 161 = 10.

This simplifies to:

3c - 161 = 10.

Next, add 161 to both sides to isolate the term with \(c\):

3c = 171.

Finally, divide both sides by 3 to solve for \(c\):

c = 57.

Thus, the final answer is 57.<|END_OF_LATENT|> The final answer is $57$.