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Let's work through the problems one by one.

### Part 1: True/False Statements
For the first part, we need to determine whether each equation is true or false by substituting the given values of \( x \).

#### 1. \( 5x - 8 = 22 \), for \( x = 6 \)
Substitute \( x = 6 \):
\[
5(6) - 8 = 30 - 8 = 22
\]
This is **true**.

#### 2. \( 4x^2 - 3(5) = 12 \), for \( x = 3 \)
Substitute \( x = 3 \):
\[
4(3^2) - 3(5) = 4(9) - 15 = 36 - 15 = 21
\]
This does **not** equal 12, so this is **false**.

#### 3. \( (8 - n)^2 + 13 = 23 \), for \( n = 3 \)
Substitute \( n = 3 \):
\[
(8 - 3)^2 + 13 = 5^2 + 13 = 25 + 13 = 38
\]
This does **not** equal 23, so this is **false**.

#### 4. \( 17x - 8(2x - 4) = 32 \), for \( x = 3 \)
Substitute \( x = 3 \):
\[
17(3) - 8(2(3) - 4) = 51 - 8(6 - 4) = 51 - 8(2) = 51 - 16 = 35
\]
This does **not** equal 32, so this is **false**.

#### 5. \( 2x + 12 + 8x \geq 32 \), for \( x = 2 \)
Substitute \( x = 2 \):
\[
2(2) + 12 + 8(2) \geq 32 \implies 4 + 12 + 16 = 32
\]
Since \( 32 \geq 32 \), this is **true**.

#### 6. \( 6(3x - 2) + 5 < 50 \), for \( x = 3 \)
Substitute \( x = 3 \):
\[
6(3(3) - 2) + 5 < 50 \implies 6(9 - 2) + 5 < 50 \implies 6(7) + 5 = 42 + 5 = 47
\]
Since \( 47 < 50 \), this is **true**.

---

### Part 2: Solution Set
For the second part, we need to find which values of \( x \) from the given replacement sets satisfy the equation or inequality.

#### 7. \( 5x + 2 = 17 \), Replacement Set: \( \{1, 2, 3, 4\} \)
Solve for \( x \):
\[
5x + 2 = 17 \implies 5x = 15 \implies x = 3
\]
The solution set is \( \{3\} \).

#### 8. \( 3x - 2 > 4 \), Replacement Set: \( \{2, 3, 4, 5\} \)
Solve for \( x \):
\[
3x - 2 > 4 \implies 3x > 6 \implies x > 2
\]
The solution set is \( \{3, 4, 5\} \).

#### 9. \( 2x^2 + 4 = 54 \), Replacement Set: \( \{1, 3, 5, 7\} \)
Solve for \( x \):
\[
2x^2 + 4 = 54 \implies 2x^2 = 50 \implies x^2 = 25 \implies x = \pm 5
\]
The solution set is \( \{5\} \).

#### 10. \( 7x - 7 < 30 \), Replacement Set: \( \{2, 4, 6, 8\} \)
Solve for \( x \):
\[
7x - 7 < 30 \implies 7x < 37 \implies x < \frac{37}{7} \approx 5.29
\]
The solution set is \( \{2, 4\} \) (since \( x < 5.29 \)).

#### 11. \( 2(2x + 4) \geq 20 \), Replacement Set: \( \{3, 5, 6, 9\} \)
Solve for \( x \):
\[
2(2x + 4) \geq 20 \implies 4x + 8 \geq 20 \implies 4x \geq 12 \implies x \geq 3
\]
The solution set is \( \{3, 5, 6, 9\} \).

#### 12. \( 5x - 6 = 24 \), Replacement Set: \( \{1, 2, 3, 4\} \)
Solve for \( x \):
\[
5x - 6 = 24 \implies 5x = 30 \implies x = 6
\]
No solution exists within the given set, so the solution set is \( \{\} \).

---

Would you like a more detailed explanation of any of these problems? Here are 5 related questions to expand your understanding:

1. How do you handle inequalities that involve both \( > \) and \( \geq \)?
2. What is the difference between solving an equation and solving an inequality?
3. Why do some equations not have solutions within a replacement set?
4. What happens when you square both sides of an equation?
5. How do you interpret compound inequalities?

**Tip:** When solving equations or inequalities with a replacement set, always check each value individually, especially when the set is small.

State whether each equation is true or false for the value of the variable given. Show your work to determine your answer. Find the solution set for each equation or inequality using the given replacement set for x.

Learn how to determine if equations are true or false and solve inequalities with replacement sets. This problem covers key algebraic concepts like linear equations, quadratic equations, and inequalities. Suitable for students in Grades 7-9.

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