Set of four algebra problems focused on solving compound inequalities using 'and' and 'or' logic.

Let's solve each inequality algebraically, then graph the result, and finally write the answer in interval notation.

---

A) $ x - 2 < 5 $ or $ 3x \geq 27 $

#### Step 1: Solve each inequality separately

1. $ x - 2 < 5 $

Add 2 to both sides:

$$
x < 7
$$

2. $ 3x \geq 27 $

Divide both sides by 3:

$$
x \geq 9
$$

#### Step 2: Combine using "or"

We are looking for values of $ x $ that satisfy either condition.

So, the solution is:
$$
x < 7 \quad \text{or} \quad x \geq 9
$$

#### Step 3: Interval Notation

$$
(-\infty, 7) \cup [9, \infty)
$$

#### Step 4: Graph

On a number line:
- Open circle at 7, shade left (all numbers less than 7).
- Closed circle at 9, shade right (all numbers greater than or equal to 9).

```
<----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|---->
-∞ 0 1 2 3 4 5 6 7 8 9 10 ∞
o================= | ====================>
```

(Note: The open circle at 7, closed at 9, with shading on both sides.)

---

B) $ -5 \leq x + 3 $ and $ x + 3 \leq 9 $

This is a compound inequality connected by "and", so we solve both and find the intersection.

#### Step 1: Solve each inequality

1. $ -5 \leq x + 3 $

Subtract 3:

$$
-8 \leq x
$$

2. $ x + 3 \leq 9 $

Subtract 3:

$$
x \leq 6
$$

#### Step 2: Combine using "and"

So,
$$
-8 \leq x \leq 6
$$

#### Step 3: Interval Notation

$$
[-8, 6]
$$

#### Step 4: Graph

Closed circles at -8 and 6, shaded between them.

```
<----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|---->
-10 -8 -6 -4 -2 0 2 4 6 8 10 12 ∞
[=====================]
```

---

C) $ 4x + 1 \leq -11 $ or $ \frac{x}{2} - 5 > -3 $

Solve each inequality separately.

#### 1. $ 4x + 1 \leq -11 $

Subtract 1:

$$
4x \leq -12
$$

Divide by 4:

$$
x \leq -3
$$

#### 2. $ \frac{x}{2} - 5 > -3 $

Add 5:

$$
\frac{x}{2} > 2
$$

Multiply both sides by 2:

$$
x > 4
$$

#### Combine using "or"

So, either $ x \leq -3 $ or $ x > 4 $

#### Interval Notation

$$
(-\infty, -3] \cup (4, \infty)
$$

#### Graph

- Closed circle at -3, shade left.
- Open circle at 4, shade right.

```
<----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|---->
-∞ -5 -4 -3 -2 -1 0 1 2 3 4 5 6 ∞
[================== | ====================>
```

(Shading from -∞ to -3 (inclusive), and from 4 (exclusive) to ∞)

---

D) $ -6 < \frac{x - 12}{4} < -2 $

This is a compound inequality — one expression between two values.

We can solve it all at once.

#### Step 1: Multiply all parts by 4 (positive, so inequality signs stay same)

$$
-6 \cdot 4 < x - 12 < -2 \cdot 4
$$
$$
-24 < x - 12 < -8
$$

#### Step 2: Add 12 to all parts

$$
-24 + 12 < x < -8 + 12
$$
$$
-12 < x < 4
$$

#### Interval Notation

$$
(-12, 4)
$$

#### Graph

Open circles at -12 and 4, shade between.

```
<----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|---->
-14 -12 -10 -8 -6 -4 -2 0 2 4 6 8 10 ∞
o============================o
```

---

✔ Final Answers:

| Part | Solution (Algebraic) | Interval Notation | Graph Description |
|------|------------------------|-------------------|-------------------|
| A) | $ x < 7 $ or $ x \geq 9 $ | $ (-\infty, 7) \cup [9, \infty) $ | Open at 7, closed at 9, shaded left and right |
| B) | $ -8 \leq x \leq 6 $ | $ [-8, 6] $ | Closed at both ends, shaded between |
| C) | $ x \leq -3 $ or $ x > 4 $ | $ (-\infty, -3] \cup (4, \infty) $ | Closed at -3, open at 4, shaded left and right |
| D) | $ -12 < x < 4 $ | $ (-12, 4) $ | Open at both ends, shaded between |

Let me know if you'd like visual graphs drawn in text or as images!