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The given inequality is a compound inequality:

\[
3y < 5 - 2y < 7 + y
\]

To solve this, we will break it into two separate inequalities:

1. \(3y < 5 - 2y\)
2. \(5 - 2y < 7 + y\)

### Step 1: Solve \(3y < 5 - 2y\)
Add \(2y\) to both sides to combine the \(y\)-terms:
\[
3y + 2y < 5
\]
\[
5y < 5
\]
Now divide by 5:
\[
y < 1
\]

### Step 2: Solve \(5 - 2y < 7 + y\)
First, subtract \(5\) from both sides:
\[
-2y < 7 + y - 5
\]
\[
-2y < 2 + y
\]
Now subtract \(y\) from both sides:
\[
-2y - y < 2
\]
\[
-3y < 2
\]
Divide by \(-3\) (remember to flip the inequality sign):
\[
y > -\frac{2}{3}
\]

### Final Solution:
We now combine the two results:
\[
-\frac{2}{3} < y < 1
\]

So, the solution is \(y\) lies between \(-\frac{2}{3}\) and \(1\), written as:

\[
y \in \left( -\frac{2}{3}, 1 \right)
\]

Would you like more details or clarifications? 

Here are 5 related questions to explore further:
1. How do we handle compound inequalities involving absolute values?
2. Can you explain how to graph the solution on a number line?
3. What happens if the inequality contains fractions?
4. How do we solve compound inequalities involving three parts like the one above?
5. How does flipping the inequality work when multiplying or dividing by a negative number?

**Tip:** Always remember to flip the inequality sign when multiplying or dividing by a negative number!

Learn how to solve the compound inequality 3y < 5 - 2y < 7 + y with a step-by-step solution. This problem involves key algebraic concepts like combining like terms and handling inequalities. Suitable for students in Grades 8-10.

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