Algebraic equations practice problems for solving linear equations with one variable.

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: \( 2(x + 5) = 16 \)

1. Distribute the 2 on the left-hand side:
\[
2x + 10 = 16
\]

2. Subtract 10 from both sides to isolate the term with \( x \):
\[
2x = 6
\]

3. Divide both sides by 2:
\[
x = 3
\]

Solution: \( x = 3 \)

---

Equation 2: \( 3(t + 1) = 18 \)

1. Distribute the 3 on the left-hand side:
\[
3t + 3 = 18
\]

2. Subtract 3 from both sides to isolate the term with \( t \):
\[
3t = 15
\]

3. Divide both sides by 3:
\[
t = 5
\]

Solution: \( t = 5 \)

---

Equation 3: \( 2(3y - 5) = 14 \)

1. Distribute the 2 on the left-hand side:
\[
6y - 10 = 14
\]

2. Add 10 to both sides to isolate the term with \( y \):
\[
6y = 24
\]

3. Divide both sides by 6:
\[
y = 4
\]

Solution: \( y = 4 \)

---

Equation 4: \( 4(3t - 2) = 88 \)

1. Distribute the 4 on the left-hand side:
\[
12t - 8 = 88
\]

2. Add 8 to both sides to isolate the term with \( t \):
\[
12t = 96
\]

3. Divide both sides by 12:
\[
t = 8
\]

Solution: \( t = 8 \)

---

Equation 5: \( 2(3x + 1) = 11 \)

1. Distribute the 2 on the left-hand side:
\[
6x + 2 = 11
\]

2. Subtract 2 from both sides to isolate the term with \( x \):
\[
6x = 9
\]

3. Divide both sides by 6:
\[
x = \frac{9}{6} = \frac{3}{2}
\]

Solution: \( x = \frac{3}{2} \)

---

Equation 6: \( 6(3k + 5) = 39 \)

1. Distribute the 6 on the left-hand side:
\[
18k + 30 = 39
\]

2. Subtract 30 from both sides to isolate the term with \( k \):
\[
18k = 9
\]

3. Divide both sides by 18:
\[
k = \frac{9}{18} = \frac{1}{2}
\]

Solution: \( k = \frac{1}{2} \)

---

Equation 7: \( 9(3x - 5) = 9 \)

1. Distribute the 9 on the left-hand side:
\[
27x - 45 = 9
\]

2. Add 45 to both sides to isolate the term with \( x \):
\[
27x = 54
\]

3. Divide both sides by 27:
\[
x = \frac{54}{27} = 2
\]

Solution: \( x = 2 \)

---

Equation 8: \( 3(t + 7) = 15 \)

1. Distribute the 3 on the left-hand side:
\[
3t + 21 = 15
\]

2. Subtract 21 from both sides to isolate the term with \( t \):
\[
3t = -6
\]

3. Divide both sides by 3:
\[
t = -2
\]

Solution: \( t = -2 \)

---

Equation 9: \( 5y + 4 = 3y + 6 \)

1. Subtract \( 3y \) from both sides to get all \( y \)-terms on one side:
\[
5y - 3y + 4 = 6
\]
\[
2y + 4 = 6
\]

2. Subtract 4 from both sides to isolate the term with \( y \):
\[
2y = 2
\]

3. Divide both sides by 2:
\[
y = 1
\]

Solution: \( y = 1 \)

---

Equation 10: \( 5t + 3 = 2t + 15 \)

1. Subtract \( 2t \) from both sides to get all \( t \)-terms on one side:
\[
5t - 2t + 3 = 15
\]
\[
3t + 3 = 15
\]

2. Subtract 3 from both sides to isolate the term with \( t \):
\[
3t = 12
\]

3. Divide both sides by 3:
\[
t = 4
\]

Solution: \( t = 4 \)

---

Equation 11: \( 6k + 5 = 2k + 1 \)

1. Subtract \( 2k \) from both sides to get all \( k \)-terms on one side:
\[
6k - 2k + 5 = 1
\]
\[
4k + 5 = 1
\]

2. Subtract 5 from both sides to isolate the term with \( k \):
\[
4k = -4
\]

3. Divide both sides by 4:
\[
k = -1
\]

Solution: \( k = -1 \)

---

Equation 12: \( 8s - 1 = 6s - 5 \)

1. Subtract \( 6s \) from both sides to get all \( s \)-terms on one side:
\[
8s - 6s - 1 = -5
\]
\[
2s - 1 = -5
\]

2. Add 1 to both sides to isolate the term with \( s \):
\[
2s = -4
\]

3. Divide both sides by 2:
\[
s = -2
\]

Solution: \( s = -2 \)

---

Equation 13: \( 5(x - 2) = 3(x + 4) \)

1. Distribute on both sides:
\[
5x - 10 = 3x + 12
\]

2. Subtract \( 3x \) from both sides to get all \( x \)-terms on one side:
\[
5x - 3x - 10 = 12
\]
\[
2x - 10 = 12
\]

3. Add 10 to both sides to isolate the term with \( x \):
\[
2x = 22
\]

4. Divide both sides by 2:
\[
x = 11
\]

Solution: \( x = 11 \)

---

Equation 14: \( 3(h - 6) = 2(5 - 2h) \)

1. Distribute on both sides:
\[
3h - 18 = 10 - 4h
\]

2. Add \( 4h \) to both sides to get all \( h \)-terms on one side:
\[
3h + 4h - 18 = 10
\]
\[
7h - 18 = 10
\]

3. Add 18 to both sides to isolate the term with \( h \):
\[
7h = 28
\]

4. Divide both sides by 7:
\[
h = 4
\]

Solution: \( h = 4 \)

---

Equation 15: \( 2(5c + 2) - 2c = 3(2c + 3) + 7 \)

1. Distribute on both sides:
\[
10c + 4 - 2c = 6c + 9 + 7
\]

2. Simplify both sides:
\[
8c + 4 = 6c + 16
\]

3. Subtract \( 6c \) from both sides to get all \( c \)-terms on one side:
\[
8c - 6c + 4 = 16
\]
\[
2c + 4 = 16
\]

4. Subtract 4 from both sides to isolate the term with \( c \):
\[
2c = 12
\]

5. Divide both sides by 2:
\[
c = 6
\]

Solution: \( c = 6 \)

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ x = 3 \\
2. & \ t = 5 \\
3. & \ y = 4 \\
4. & \ t = 8 \\
5. & \ x = \frac{3}{2} \\
6. & \ k = \frac{1}{2} \\
7. & \ x = 2 \\
8. & \ t = -2 \\
9. & \ y = 1 \\
10. & \ t = 4 \\
11. & \ k = -1 \\
12. & \ s = -2 \\
13. & \ x = 11 \\
14. & \ h = 4 \\
15. & \ c = 6 \\
\end{aligned}
}
\]