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Let's go through the set of equations one by one and solve them:

### 1. \( x + 5 = 7 \)
To solve for \( x \):
\[
x = 7 - 5 = 2
\]

### 2. \( 2 + x = 7 \)
To solve for \( x \):
\[
x = 7 - 2 = 5
\]

### 3. \( x + 8 = 6 \)
To solve for \( x \):
\[
x = 6 - 8 = -2
\]

### 4. \( x - 3 = 12 \)
To solve for \( x \):
\[
x = 12 + 3 = 15
\]

### 5. \( x + 15 = 20 \)
To solve for \( x \):
\[
x = 20 - 15 = 5
\]

### 6. \( \frac{x}{3} = 4 \)
To solve for \( x \):
\[
x = 4 \times 3 = 12
\]

### 7. \( 3x - 12 = 0 \)
To solve for \( x \):
\[
3x = 12 \quad \Rightarrow \quad x = \frac{12}{3} = 4
\]

### 8. \( 2x - 3 = 5 + x \)
To solve for \( x \):
\[
2x - x = 5 + 3 \quad \Rightarrow \quad x = 8
\]

### 9. \( 8x - 3 = 4 + 7x \)
To solve for \( x \):
\[
8x - 7x = 4 + 3 \quad \Rightarrow \quad x = 7
\]

### 10. \( \frac{x + 3}{3} + \frac{x}{3} - x = 2 \)
Combine the terms on the left:
\[
\frac{(x + 3) + x}{3} - x = 2 \quad \Rightarrow \quad \frac{2x + 3}{3} - x = 2
\]
Multiply everything by 3 to eliminate the denominator:
\[
2x + 3 - 3x = 6 \quad \Rightarrow \quad -x + 3 = 6 \quad \Rightarrow \quad -x = 6 - 3 = 3
\]
So:
\[
x = -3
\]

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All equations are now solved. Do you want any further details or explanations on a specific step?

Here are 5 related questions to test your understanding:
1. How would you solve an equation with fractions on both sides?
2. How do you check if a solution to an equation is correct?
3. Can an equation have more than one solution?
4. What are the steps to solve a quadratic equation?
5. How do you solve equations with variables on both sides?

**Tip**: Always check your solution by substituting it back into the original equation to verify it works.

Solve the set of algebraic equations (10 in total).

Learn how to solve a variety of linear equations, including equations with variables on both sides and equations with fractions. This problem covers essential algebraic techniques and is suitable for students in Grades 6-8.

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