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?

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.