The given equation is:

\[
(n+1)a + nb = 0
\]

To solve for \( a \) in terms of \( b \) and \( n \), we can follow these steps:

### Step 1: Expand and rearrange the equation
\[
(n+1)a + nb = 0
\]

This can be rewritten as:
\[
a(n+1) = -nb
\]

### Step 2: Solve for \( a \)
To isolate \( a \), divide both sides of the equation by \( (n+1) \) (assuming \( n \neq -1 \), so that \( n+1 \neq 0 \)):

\[
a = \frac{-nb}{n+1}
\]

Thus, the solution for \( a \) in terms of \( b \) and \( n \) is:

\[
a = \frac{-nb}{n+1}
\]

Would you like further details or explanations?

Here are 5 related questions for further exploration:

1. What happens if \( n = -1 \) in this equation?
2. Can you solve for \( b \) in terms of \( a \) and \( n \)?
3. How would you interpret this equation geometrically?
4. What happens to the value of \( a \) as \( n \) approaches infinity?
5. How would you solve the system if you had two similar equations?

**Tip:** Always check for values of variables that may cause division by zero, like \( n = -1 \) in this case!

Learn how to solve the linear equation (n+1)a + nb = 0 with step-by-step guidance. This solution demonstrates how to express a in terms of b and n by applying basic algebraic manipulation. Suitable for high school students in Grades 9-12.

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