Find the value of $n$ that satisfies $\frac{1}{n+1} + \frac{2}{n+1} + \frac{n}{n+1} = 3$.
Combining the fractions on the left gives $\dfrac{n+3}{n+1} = 3$.  Multiplying both sides by $n+1$ gives $n+3 = 3(n+1)$.  Expanding the right side gives $n+3 = 3n+3$.  Subtracting $n$ and 3 from both sides gives $0=2n$, so $n=\boxed{0}$.
0
Level 2
Algebra
Problem: Find the value of $n$ that satisfies $\frac{1}{n+1} + \frac{2}{n+1} + \frac{n}{n+1} = 3$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation, first combine the fractions on the left side. The common denominator is \( n + 1 \). Thus, we can rewrite the equation as:

\[
\frac{1 + 2 + n}{n + 1} = 3
\]

This simplifies to:

\[
\frac{n + 3}{n + 1} = 3
\]

Next, cross-multiply to eliminate the fraction:

\[
n + 3 = 3(n + 1)
\]

Expanding the right side gives:

\[
n + 3 = 3n + 3
\]

Now, isolate \( n \) by subtracting \( n \) from both sides:

\[
3 = 2n + 3
\]

Subtract 3 from both sides:

\[
0 = 2n
\]

Dividing both sides by 2 yields:

\[
n = 0
\]

Thus, the value of \( n \) that satisfies the original equation is \( 0 \).<|END_OF_LATENT|> The final answer is $0$.