Printable worksheet featuring ten algebra problems focused on solving linear equations and inequalities for students to practice.

Problem: Solve the linear equations and inequalities in one variable.

The worksheet contains 10 problems. Below, I will solve each problem step by step.

---

#### Problem 1:
$$
15p - 5p - 4 = 8
$$

1. Combine like terms on the left-hand side:
$$
15p - 5p = 10p
$$
So the equation becomes:
$$
10p - 4 = 8
$$

2. Add 4 to both sides to isolate the term with \( p \):
$$
10p - 4 + 4 = 8 + 4
$$
$$
10p = 12
$$

3. Divide both sides by 10 to solve for \( p \):
$$
p = \frac{12}{10} = \frac{6}{5}
$$

Solution:
$$
\boxed{\frac{6}{5}}
$$

---

#### Problem 2:
$$
8b = 2b + 6 + 4
$$

1. Simplify the right-hand side:
$$
2b + 6 + 4 = 2b + 10
$$
So the equation becomes:
$$
8b = 2b + 10
$$

2. Subtract \( 2b \) from both sides to isolate the term with \( b \):
$$
8b - 2b = 2b + 10 - 2b
$$
$$
6b = 10
$$

3. Divide both sides by 6 to solve for \( b \):
$$
b = \frac{10}{6} = \frac{5}{3}
$$

Solution:
$$
\boxed{\frac{5}{3}}
$$

---

#### Problem 3:
$$
7a - 4a - 2 = 5
$$

1. Combine like terms on the left-hand side:
$$
7a - 4a = 3a
$$
So the equation becomes:
$$
3a - 2 = 5
$$

2. Add 2 to both sides to isolate the term with \( a \):
$$
3a - 2 + 2 = 5 + 2
$$
$$
3a = 7
$$

3. Divide both sides by 3 to solve for \( a \):
$$
a = \frac{7}{3}
$$

Solution:
$$
\boxed{\frac{7}{3}}
$$

---

#### Problem 4:
$$
9x - 6x - 8 = 2
$$

1. Combine like terms on the left-hand side:
$$
9x - 6x = 3x
$$
So the equation becomes:
$$
3x - 8 = 2
$$

2. Add 8 to both sides to isolate the term with \( x \):
$$
3x - 8 + 8 = 2 + 8
$$
$$
3x = 10
$$

3. Divide both sides by 3 to solve for \( x \):
$$
x = \frac{10}{3}
$$

Solution:
$$
\boxed{\frac{10}{3}}
$$

---

#### Problem 5:
$$
\frac{n}{3} = 18
$$

1. Multiply both sides by 3 to eliminate the denominator:
$$
3 \cdot \frac{n}{3} = 18 \cdot 3
$$
$$
n = 54
$$

Solution:
$$
\boxed{54}
$$

---

#### Problem 6:
$$
\frac{w}{4} - 1 = 4
$$

1. Add 1 to both sides to isolate the fraction:
$$
\frac{w}{4} - 1 + 1 = 4 + 1
$$
$$
\frac{w}{4} = 5
$$

2. Multiply both sides by 4 to eliminate the denominator:
$$
4 \cdot \frac{w}{4} = 5 \cdot 4
$$
$$
w = 20
$$

Solution:
$$
\boxed{20}
$$

---

#### Problem 7:
$$
\frac{m}{5} = 13
$$

1. Multiply both sides by 5 to eliminate the denominator:
$$
5 \cdot \frac{m}{5} = 13 \cdot 5
$$
$$
m = 65
$$

Solution:
$$
\boxed{65}
$$

---

#### Problem 8:
$$
5 = \frac{v + 14}{5}
$$

1. Multiply both sides by 5 to eliminate the denominator:
$$
5 \cdot 5 = \frac{v + 14}{5} \cdot 5
$$
$$
25 = v + 14
$$

2. Subtract 14 from both sides to isolate \( v \):
$$
25 - 14 = v + 14 - 14
$$
$$
v = 11
$$

Solution:
$$
\boxed{11}
$$

---

#### Problem 9:
$$
6 = \frac{k + 12}{k}
$$

1. Multiply both sides by \( k \) to eliminate the denominator:
$$
6 \cdot k = \frac{k + 12}{k} \cdot k
$$
$$
6k = k + 12
$$

2. Subtract \( k \) from both sides to isolate the term with \( k \):
$$
6k - k = k + 12 - k
$$
$$
5k = 12
$$

3. Divide both sides by 5 to solve for \( k \):
$$
k = \frac{12}{5}
$$

Solution:
$$
\boxed{\frac{12}{5}}
$$

---

#### Problem 10:
$$
3 > \frac{x + 8}{7}
$$

1. Multiply both sides by 7 to eliminate the denominator:
$$
3 \cdot 7 > \frac{x + 8}{7} \cdot 7
$$
$$
21 > x + 8
$$

2. Subtract 8 from both sides to isolate \( x \):
$$
21 - 8 > x + 8 - 8
$$
$$
13 > x
$$

3. Rewrite the inequality in standard form:
$$
x < 13
$$

Solution:
$$
\boxed{x < 13}
$$

---

Final Answers:

1. \(\boxed{\frac{6}{5}}\)
2. \(\boxed{\frac{5}{3}}\)
3. \(\boxed{\frac{7}{3}}\)
4. \(\boxed{\frac{10}{3}}\)
5. \(\boxed{54}\)
6. \(\boxed{20}\)
7. \(\boxed{65}\)
8. \(\boxed{11}\)
9. \(\boxed{\frac{12}{5}}\)
10. \(\boxed{x < 13}\)