Algebra equations practice problems for solving linear equations.

Problem:

The task is to solve each of the given linear equations for \( x \). The equations are listed as follows:

1. \( x - 3 = 11 \)
2. \( x - 5 = 7 \)
3. \( 2x = 10 \)
4. \( \frac{x}{6} = 2 \)
5. \( 3x = 15 \)
6. \( 4x = 12 \)
7. \( \frac{x}{4} = 10 \)
8. \( x - 7 = 13 \)
9. \( x + 6 = 15 \)
10. \( 4 + x = 9 \)
11. \( 14 = x + 5 \)

---

Solution:

#### Equation 1: \( x - 3 = 11 \)
To solve for \( x \), add 3 to both sides:
\[
x - 3 + 3 = 11 + 3
\]
\[
x = 14
\]

#### Equation 2: \( x - 5 = 7 \)
Add 5 to both sides:
\[
x - 5 + 5 = 7 + 5
\]
\[
x = 12
\]

#### Equation 3: \( 2x = 10 \)
Divide both sides by 2:
\[
\frac{2x}{2} = \frac{10}{2}
\]
\[
x = 5
\]

#### Equation 4: \( \frac{x}{6} = 2 \)
Multiply both sides by 6:
\[
6 \cdot \frac{x}{6} = 6 \cdot 2
\]
\[
x = 12
\]

#### Equation 5: \( 3x = 15 \)
Divide both sides by 3:
\[
\frac{3x}{3} = \frac{15}{3}
\]
\[
x = 5
\]

#### Equation 6: \( 4x = 12 \)
Divide both sides by 4:
\[
\frac{4x}{4} = \frac{12}{4}
\]
\[
x = 3
\]

#### Equation 7: \( \frac{x}{4} = 10 \)
Multiply both sides by 4:
\[
4 \cdot \frac{x}{4} = 4 \cdot 10
\]
\[
x = 40
\]

#### Equation 8: \( x - 7 = 13 \)
Add 7 to both sides:
\[
x - 7 + 7 = 13 + 7
\]
\[
x = 20
\]

#### Equation 9: \( x + 6 = 15 \)
Subtract 6 from both sides:
\[
x + 6 - 6 = 15 - 6
\]
\[
x = 9
\]

#### Equation 10: \( 4 + x = 9 \)
Subtract 4 from both sides:
\[
4 + x - 4 = 9 - 4
\]
\[
x = 5
\]

#### Equation 11: \( 14 = x + 5 \)
Rewrite the equation as \( x + 5 = 14 \) and subtract 5 from both sides:
\[
x + 5 - 5 = 14 - 5
\]
\[
x = 9
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ x = 14 \\
2. & \ x = 12 \\
3. & \ x = 5 \\
4. & \ x = 12 \\
5. & \ x = 5 \\
6. & \ x = 3 \\
7. & \ x = 40 \\
8. & \ x = 20 \\
9. & \ x = 9 \\
10. & \ x = 5 \\
11. & \ x = 9 \\
\end{aligned}
}
\]