Students can practice their early algebra skills by solving for the unknown variable 'y' in these simple addition equations.

Problem: Solve for the variable \( y \) in each equation.

The goal is to isolate \( y \) on one side of the equation. We will do this by performing inverse operations (such as addition or subtraction) to balance the equation.

---

#### Step-by-Step Solutions:

1. \( 14 = 5 + y \)
- Subtract 5 from both sides:
\[
14 - 5 = y
\]
\[
y = 9
\]

2. \( 13 = 5 + y \)
- Subtract 5 from both sides:
\[
13 - 5 = y
\]
\[
y = 8
\]

3. \( y + 3 = 6 \)
- Subtract 3 from both sides:
\[
y = 6 - 3
\]
\[
y = 3
\]

4. \( 10 = y + 7 \)
- Subtract 7 from both sides:
\[
10 - 7 = y
\]
\[
y = 3
\]

5. \( 12 = y + 4 \)
- Subtract 4 from both sides:
\[
12 - 4 = y
\]
\[
y = 8
\]

6. \( 8 = y + 3 \)
- Subtract 3 from both sides:
\[
8 - 3 = y
\]
\[
y = 5
\]

7. \( 9 = y + 4 \)
- Subtract 4 from both sides:
\[
9 - 4 = y
\]
\[
y = 5
\]

8. \( 14 = 9 + y \)
- Subtract 9 from both sides:
\[
14 - 9 = y
\]
\[
y = 5
\]

9. \( y + 1 = 7 \)
- Subtract 1 from both sides:
\[
y = 7 - 1
\]
\[
y = 6
\]

10. \( 3 + y = 9 \)
- Subtract 3 from both sides:
\[
y = 9 - 3
\]
\[
y = 6
\]

11. \( y + 1 = 10 \)
- Subtract 1 from both sides:
\[
y = 10 - 1
\]
\[
y = 9
\]

12. \( y + 6 = 7 \)
- Subtract 6 from both sides:
\[
y = 7 - 6
\]
\[
y = 1
\]

13. \( 1 + y = 3 \)
- Subtract 1 from both sides:
\[
y = 3 - 1
\]
\[
y = 2
\]

14. \( y + 8 = 17 \)
- Subtract 8 from both sides:
\[
y = 17 - 8
\]
\[
y = 9
\]

15. \( y + 7 = 15 \)
- Subtract 7 from both sides:
\[
y = 15 - 7
\]
\[
y = 8
\]

16. \( 4 + y = 11 \)
- Subtract 4 from both sides:
\[
y = 11 - 4
\]
\[
y = 7
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & y = 9 \\
2. & y = 8 \\
3. & y = 3 \\
4. & y = 3 \\
5. & y = 8 \\
6. & y = 5 \\
7. & y = 5 \\
8. & y = 5 \\
9. & y = 6 \\
10. & y = 6 \\
11. & y = 9 \\
12. & y = 1 \\
13. & y = 2 \\
14. & y = 9 \\
15. & y = 8 \\
16. & y = 7 \\
\end{array}
}
\]