show that the indicated implication is true.

49. |x-3|<0.5 Rightarrow |5x - 15| < 2.5

|x+2|<0.3 Rightarrow |4x + 8| < 1.2

To show that each implication is true, let's analyze them one at a time.

### Problem 1: \( |x - 3| < 0.5 \Rightarrow |5x - 15| < 2.5 \)

1. **Start with the assumption:** We assume \( |x - 3| < 0.5 \) and aim to show that \( |5x - 15| < 2.5 \).
2. **Rewrite \( |5x - 15| \):**
   Notice that:
   \[
   |5x - 15| = |5(x - 3)|
   \]
3. **Apply the property of absolute values:** We can factor out the 5:
   \[
   |5(x - 3)| = 5 |x - 3|
   \]
4. **Substitute from the assumption:** Since \( |x - 3| < 0.5 \), we get
   \[
   5 |x - 3| < 5 \cdot 0.5 = 2.5
   \]
5. **Conclusion:** This implies \( |5x - 15| < 2.5 \), as required.

Thus, the first implication is true.

---

### Problem 2: \( |x + 2| < 0.3 \Rightarrow |4x + 8| < 1.2 \)

1. **Start with the assumption:** We assume \( |x + 2| < 0.3 \) and aim to show that \( |4x + 8| < 1.2 \).
2. **Rewrite \( |4x + 8| \):**
   Notice that:
   \[
   |4x + 8| = |4(x + 2)|
   \]
3. **Apply the property of absolute values:** We can factor out the 4:
   \[
   |4(x + 2)| = 4 |x + 2|
   \]
4. **Substitute from the assumption:** Since \( |x + 2| < 0.3 \), we get
   \[
   4 |x + 2| < 4 \cdot 0.3 = 1.2
   \]
5. **Conclusion:** This implies \( |4x + 8| < 1.2 \), as required.

Thus, the second implication is also true.

---

Both implications are valid.

---

Do you need further details or have any questions?

### Related Questions

1. How do absolute value inequalities behave under scalar multiplication?
2. What are some other common properties of absolute values used in proofs?
3. Can these inequalities be solved by considering the ranges of \( x \) explicitly?
4. How do absolute value inequalities relate to distance on the real number line?
5. Could this problem be generalized for other linear expressions?

**Tip:** In inequalities involving absolute values, always consider factoring constants out to simplify the expressions.

Show that the indicated implication is true. 

49. |x-3| < 0.5 Rightarrow |5x - 15| < 2.5 

|x+2| < 0.3 Rightarrow |4x + 8| < 1.2

Explore step-by-step solutions for proving the implications |x-3| < 0.5 ⇒ |5x - 15| < 2.5 and |x+2| < 0.3 ⇒ |4x + 8| < 1.2. This solution covers absolute value inequalities, scalar multiplication, and properties of absolute values, suitable for Grades 9-11.

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