= 
y 
2
 −4
y+3−2y−1
 what happened to the -2y?

I see where the confusion might be coming from. Let's carefully go through the simplification of the expression again.

Given expression:

\[
\frac{|y + 3| - |2y + 1|}{y^2 - 4}
\]

### Step-by-Step Simplification

1. **Rewrite the Absolute Values:**
   - As \( y \) approaches 2:
     - \( |y + 3| = y + 3 \) (since \( y + 3 \) is positive near 2)
     - \( |2y + 1| = 2y + 1 \) (since \( 2y + 1 \) is also positive near 2)

   So, the expression inside the absolute values becomes:
   \[
   \frac{(y + 3) - (2y + 1)}{y^2 - 4}
   \]

2. **Simplify the Numerator:**
   - Distribute the negative sign in the numerator:
     \[
     y + 3 - 2y - 1
     \]
   - Combine like terms:
     \[
     (y - 2y) + (3 - 1) = -y + 2
     \]
   - The numerator simplifies to:
     \[
     \frac{-y + 2}{y^2 - 4}
     \]

3. **Factor the Denominator:**
   - Factor \( y^2 - 4 \) as a difference of squares:
     \[
     y^2 - 4 = (y - 2)(y + 2)
     \]

4. **Combine the Expression:**
   - The entire expression now is:
     \[
     \frac{-y + 2}{(y - 2)(y + 2)}
     \]

5. **Simplify Further:**
   - Notice that \( -y + 2 \) can be rewritten as \( -(y - 2) \). This step is crucial:
     \[
     \frac{-(y - 2)}{(y - 2)(y + 2)}
     \]
   - Now, cancel out the \( y - 2 \) term from the numerator and denominator (since \( y \) is approaching 2 and \( y - 2 \) is not zero):
     \[
     \frac{-1}{y + 2}
     \]

6. **Evaluate the Limit:**
   - Substitute \( y = 2 \):
     \[
     \frac{-1}{2 + 2} = \frac{-1}{4}
     \]

Thus, the correct simplification of the expression keeps the \( -2y \) term, and it simplifies properly to give the final answer \( \frac{-1}{4} \).

Would you like any further clarification or details on any part of the process?

---

Here are 5 related questions you might consider:

1. Why is it important to correctly factor and simplify expressions when solving limits?
2. How does one handle limits when the expression is indeterminate?
3. What are some strategies for dealing with absolute values in limit problems?
4. Can limits always be evaluated by direct substitution after simplification?
5. How does the limit change if a different value of \( y \) is approached, such as \( y \to -2 \)?

**Tip:** When simplifying algebraic expressions, always double-check the combination of like terms to avoid mistakes.

Explore the concept of absolute value and limits with a detailed step-by-step explanation. Learn how to simplify expressions involving absolute values and evaluate limits. Suitable for advanced high school students and those preparing for calculus.

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