Solve the following algebraic equations in the Beta Exercise worksheet.

To solve the given equations, we will go through each one step by step. Let's start with the first equation and proceed systematically.

---

Equation 1:

\[
\frac{2}{3}x + 4 = \frac{1}{6}x
\]

#### Step 1: Eliminate the fractions by finding a common denominator.
The least common denominator (LCD) of 3 and 6 is 6. Multiply every term by 6:
\[
6 \cdot \left(\frac{2}{3}x\right) + 6 \cdot 4 = 6 \cdot \left(\frac{1}{6}x\right)
\]
\[
4x + 24 = x
\]

#### Step 2: Isolate \( x \).
Subtract \( x \) from both sides:
\[
4x - x + 24 = 0
\]
\[
3x + 24 = 0
\]

Subtract 24 from both sides:
\[
3x = -24
\]

Divide by 3:
\[
x = -8
\]

#### Solution for Equation 1:
\[
\boxed{-8}
\]

---

Equation 2:

\[
\frac{2x + 4}{3} = \frac{x}{6}
\]

#### Step 1: Eliminate the fractions by finding a common denominator.
The LCD of 3 and 6 is 6. Multiply every term by 6:
\[
6 \cdot \left(\frac{2x + 4}{3}\right) = 6 \cdot \left(\frac{x}{6}\right)
\]
\[
2(2x + 4) = x
\]

#### Step 2: Simplify the left-hand side.
\[
4x + 8 = x
\]

#### Step 3: Isolate \( x \).
Subtract \( x \) from both sides:
\[
4x - x + 8 = 0
\]
\[
3x + 8 = 0
\]

Subtract 8 from both sides:
\[
3x = -8
\]

Divide by 3:
\[
x = -\frac{8}{3}
\]

#### Solution for Equation 2:
\[
\boxed{-\frac{8}{3}}
\]

---

Equation 3:

\[
\frac{2x + 4}{3} = \frac{x}{9}
\]

#### Step 1: Eliminate the fractions by finding a common denominator.
The LCD of 3 and 9 is 9. Multiply every term by 9:
\[
9 \cdot \left(\frac{2x + 4}{3}\right) = 9 \cdot \left(\frac{x}{9}\right)
\]
\[
3(2x + 4) = x
\]

#### Step 2: Simplify the left-hand side.
\[
6x + 12 = x
\]

#### Step 3: Isolate \( x \).
Subtract \( x \) from both sides:
\[
6x - x + 12 = 0
\]
\[
5x + 12 = 0
\]

Subtract 12 from both sides:
\[
5x = -12
\]

Divide by 5:
\[
x = -\frac{12}{5}
\]

#### Solution for Equation 3:
\[
\boxed{-\frac{12}{5}}
\]

---

Equation 4:

\[
\frac{2x + 4}{6} = \frac{x}{9}
\]

#### Step 1: Eliminate the fractions by finding a common denominator.
The LCD of 6 and 9 is 18. Multiply every term by 18:
\[
18 \cdot \left(\frac{2x + 4}{6}\right) = 18 \cdot \left(\frac{x}{9}\right)
\]
\[
3(2x + 4) = 2x
\]

#### Step 2: Simplify the left-hand side.
\[
6x + 12 = 2x
\]

#### Step 3: Isolate \( x \).
Subtract \( 2x \) from both sides:
\[
6x - 2x + 12 = 0
\]
\[
4x + 12 = 0
\]

Subtract 12 from both sides:
\[
4x = -12
\]

Divide by 4:
\[
x = -3
\]

#### Solution for Equation 4:
\[
\boxed{-3}
\]

---

Equation 5:

\[
\frac{2(x + 4)}{6} = \frac{x}{9}
\]

#### Step 1: Simplify the left-hand side.
\[
\frac{2(x + 4)}{6} = \frac{x + 4}{3}
\]
So the equation becomes:
\[
\frac{x + 4}{3} = \frac{x}{9}
\]

#### Step 2: Eliminate the fractions by finding a common denominator.
The LCD of 3 and 9 is 9. Multiply every term by 9:
\[
9 \cdot \left(\frac{x + 4}{3}\right) = 9 \cdot \left(\frac{x}{9}\right)
\]
\[
3(x + 4) = x
\]

#### Step 3: Simplify the left-hand side.
\[
3x + 12 = x
\]

#### Step 4: Isolate \( x \).
Subtract \( x \) from both sides:
\[
3x - x + 12 = 0
\]
\[
2x + 12 = 0
\]

Subtract 12 from both sides:
\[
2x = -12
\]

Divide by 2:
\[
x = -6
\]

#### Solution for Equation 5:
\[
\boxed{-6}
\]

---

Equation 6:

\[
\frac{2(x + 4)}{6} = \frac{x + 1}{9}
\]

#### Step 1: Simplify the left-hand side.
\[
\frac{2(x + 4)}{6} = \frac{x + 4}{3}
\]
So the equation becomes:
\[
\frac{x + 4}{3} = \frac{x + 1}{9}
\]

#### Step 2: Eliminate the fractions by finding a common denominator.
The LCD of 3 and 9 is 9. Multiply every term by 9:
\[
9 \cdot \left(\frac{x + 4}{3}\right) = 9 \cdot \left(\frac{x + 1}{9}\right)
\]
\[
3(x + 4) = x + 1
\]

#### Step 3: Simplify the left-hand side.
\[
3x + 12 = x + 1
\]

#### Step 4: Isolate \( x \).
Subtract \( x \) from both sides:
\[
3x - x + 12 = 1
\]
\[
2x + 12 = 1
\]

Subtract 12 from both sides:
\[
2x = -11
\]

Divide by 2:
\[
x = -\frac{11}{2}
\]

#### Solution for Equation 6:
\[
\boxed{-\frac{11}{2}}
\]

---

Equation 7:

\[
\frac{2(x + 4)}{6} = \frac{x}{5}
\]

#### Step 1: Simplify the left-hand side.
\[
\frac{2(x + 4)}{6} = \frac{x + 4}{3}
\]
So the equation becomes:
\[
\frac{x + 4}{3} = \frac{x}{5}
\]

#### Step 2: Eliminate the fractions by finding a common denominator.
The LCD of 3 and 5 is 15. Multiply every term by 15:
\[
15 \cdot \left(\frac{x + 4}{3}\right) = 15 \cdot \left(\frac{x}{5}\right)
\]
\[
5(x + 4) = 3x
\]

#### Step 3: Simplify the left-hand side.
\[
5x + 20 = 3x
\]

#### Step 4: Isolate \( x \).
Subtract \( 3x \) from both sides:
\[
5x - 3x + 20 = 0
\]
\[
2x + 20 = 0
\]

Subtract 20 from both sides:
\[
2x = -20
\]

Divide by 2:
\[
x = -10
\]

#### Solution for Equation 7:
\[
\boxed{-10}
\]

---

Equation 8:

\[
\frac{2(x + 4) + 1}{6} = \frac{x}{5} - 2
\]

#### Step 1: Simplify the left-hand side.
\[
\frac{2(x + 4) + 1}{6} = \frac{2x + 8 + 1}{6} = \frac{2x + 9}{6}
\]
So the equation becomes:
\[
\frac{2x + 9}{6} = \frac{x}{5} - 2
\]

#### Step 2: Eliminate the fractions by finding a common denominator.
The LCD of 6 and 5 is 30. Multiply every term by 30:
\[
30 \cdot \left(\frac{2x + 9}{6}\right) = 30 \cdot \left(\frac{x}{5}\right) - 30 \cdot 2
\]
\[
5(2x + 9) = 6x - 60
\]

#### Step 3: Simplify the left-hand side.
\[
10x + 45 = 6x - 60
\]

#### Step 4: Isolate \( x \).
Subtract \( 6x \) from both sides:
\[
10x - 6x + 45 = -60
\]
\[
4x + 45 = -60
\]

Subtract 45 from both sides:
\[
4x = -105
\]

Divide by 4:
\[
x = -\frac{105}{4}
\]

#### Solution for Equation 8:
\[
\boxed{-\frac{105}{4}}
\]

---

Equation 9:

\[
\frac{2(x + 4)}{6} + 1 = \frac{x}{5} - 2
\]

#### Step 1: Simplify the left-hand side.
\[
\frac{2(x + 4)}{6} = \frac{x + 4}{3}
\]
So the equation becomes:
\[
\frac{x + 4}{3} + 1 = \frac{x}{5} - 2
\]

#### Step 2: Eliminate the fractions by finding a common denominator.
The LCD of 3 and 5 is 15. Multiply every term by 15:
\[
15 \cdot \left(\frac{x + 4}{3}\right) + 15 \cdot 1 = 15 \cdot \left(\frac{x}{5}\right) - 15 \cdot 2
\]
\[
5(x + 4) + 15 = 3x - 30
\]

#### Step 3: Simplify the left-hand side.
\[
5x + 20 + 15 = 3x - 30
\]
\[
5x + 35 = 3x - 30
\]

#### Step 4: Isolate \( x \).
Subtract \( 3x \) from both sides:
\[
5x - 3x + 35 = -30
\]
\[
2x + 35 = -30
\]

Subtract 35 from both sides:
\[
2x = -65
\]

Divide by 2:
\[
x = -\frac{65}{2}
\]

#### Solution for Equation 9:
\[
\boxed{-\frac{65}{2}}
\]

---

Final Answers:

\[
\boxed{-8, -\frac{8}{3}, -\frac{12}{5}, -3, -6, -\frac{11}{2}, -10, -\frac{105}{4}, -\frac{65}{2}}
\]