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Let's solve the given equation step-by-step. The equation is:

\[
\frac{3(y + 3)}{y + 1} + 2 = \frac{3y + 1}{y + 1}
\]

### Step 1: Simplify the left-hand side

First, expand the numerator of the first term on the left-hand side:

\[
3(y + 3) = 3y + 9
\]

Now the equation becomes:

\[
\frac{3y + 9}{y + 1} + 2 = \frac{3y + 1}{y + 1}
\]

### Step 2: Eliminate the denominators

To eliminate the fractions, multiply both sides of the equation by \( y + 1 \) (assuming \( y + 1 \neq 0 \)):

\[
(3y + 9) + 2(y + 1) = 3y + 1
\]

### Step 3: Distribute the 2 on the left-hand side

\[
(3y + 9) + 2y + 2 = 3y + 1
\]

Simplifying the left-hand side:

\[
3y + 9 + 2y + 2 = 3y + 1
\]

\[
5y + 11 = 3y + 1
\]

### Step 4: Solve for \( y \)

Move all the terms involving \( y \) to one side and constants to the other side:

\[
5y - 3y = 1 - 11
\]

\[
2y = -10
\]

\[
y = -5
\]

### Step 5: Verify the solution

We should check that \( y = -5 \) does not make any denominator zero. The denominators are \( y + 1 \), so for \( y = -5 \), \( y + 1 = -4 \), which is non-zero. Thus, the solution is valid.

### Final Answer:
The solution to the equation is:

\[
y = -5
\]

Would you like a deeper explanation of any steps, or do you have other questions?

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Here are 5 related questions to extend your understanding:
1. How would the solution change if the denominator was \( y - 1 \) instead of \( y + 1 \)?
2. What would happen if we didn't multiply both sides by \( y + 1 \) in Step 2?
3. Can you solve similar equations where multiple terms have different denominators?
4. How would you handle this equation if it included a quadratic term?
5. What is the general approach to solving rational equations?

**Tip:** Always check for any restrictions in the domain (like values that make the denominator zero) when solving rational equations.

Solve the rational equation: 3(y + 3) / (y + 1) + 2 = (3y + 1) / (y + 1)

Learn how to solve the rational equation 3(y + 3) / (y + 1) + 2 = (3y + 1) / (y + 1) with a detailed step-by-step explanation. This problem covers key algebraic concepts, such as simplifying rational expressions and solving linear equations. Suitable for students in Grades 9-12.

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