Printable math worksheet featuring 15 one-variable linear equations with fractions, designed for practice and learning.

Problem: Solve the given one-variable linear equations of the form \( \frac{x}{a} + b = c \).

We will solve each equation step by step. The general approach is:
1. Isolate the term involving \( x \) by subtracting or adding the constant term.
2. Eliminate the denominator by multiplying both sides of the equation by the denominator.
3. Solve for \( x \).

Let's solve each equation:

---

Equation 1: \( \frac{x}{3} + 3 = 6 \)

1. Subtract 3 from both sides:
\[
\frac{x}{3} + 3 - 3 = 6 - 3
\]
\[
\frac{x}{3} = 3
\]

2. Multiply both sides by 3 to eliminate the denominator:
\[
3 \cdot \frac{x}{3} = 3 \cdot 3
\]
\[
x = 9
\]

Solution: \( x = 9 \)

---

Equation 2: \( \frac{x}{7} + 3 = \frac{23}{7} \)

1. Subtract 3 from both sides. Note that 3 can be written as \( \frac{21}{7} \):
\[
\frac{x}{7} + 3 - 3 = \frac{23}{7} - \frac{21}{7}
\]
\[
\frac{x}{7} = \frac{2}{7}
\]

2. Multiply both sides by 7:
\[
7 \cdot \frac{x}{7} = 7 \cdot \frac{2}{7}
\]
\[
x = 2
\]

Solution: \( x = 2 \)

---

Equation 3: \( \frac{x}{8} + 3 = \frac{13}{4} \)

1. Subtract 3 from both sides. Note that 3 can be written as \( \frac{24}{8} \) or \( \frac{12}{4} \):
\[
\frac{x}{8} + 3 - 3 = \frac{13}{4} - \frac{12}{4}
\]
\[
\frac{x}{8} = \frac{1}{4}
\]

2. Eliminate the denominator by multiplying both sides by 8:
\[
8 \cdot \frac{x}{8} = 8 \cdot \frac{1}{4}
\]
\[
x = 2
\]

Solution: \( x = 2 \)

---

Equation 4: \( \frac{x}{5} + 8 = \frac{44}{5} \)

1. Subtract 8 from both sides. Note that 8 can be written as \( \frac{40}{5} \):
\[
\frac{x}{5} + 8 - 8 = \frac{44}{5} - \frac{40}{5}
\]
\[
\frac{x}{5} = \frac{4}{5}
\]

2. Multiply both sides by 5:
\[
5 \cdot \frac{x}{5} = 5 \cdot \frac{4}{5}
\]
\[
x = 4
\]

Solution: \( x = 4 \)

---

Equation 5: \( \frac{x}{5} - 6 = -5 \)

1. Add 6 to both sides:
\[
\frac{x}{5} - 6 + 6 = -5 + 6
\]
\[
\frac{x}{5} = 1
\]

2. Multiply both sides by 5:
\[
5 \cdot \frac{x}{5} = 5 \cdot 1
\]
\[
x = 5
\]

Solution: \( x = 5 \)

---

Equation 6: \( \frac{x}{8} + 2 = \frac{25}{8} \)

1. Subtract 2 from both sides. Note that 2 can be written as \( \frac{16}{8} \):
\[
\frac{x}{8} + 2 - 2 = \frac{25}{8} - \frac{16}{8}
\]
\[
\frac{x}{8} = \frac{9}{8}
\]

2. Multiply both sides by 8:
\[
8 \cdot \frac{x}{8} = 8 \cdot \frac{9}{8}
\]
\[
x = 9
\]

Solution: \( x = 9 \)

---

Equation 7: \( \frac{x}{7} + 6 = \frac{47}{7} \)

1. Subtract 6 from both sides. Note that 6 can be written as \( \frac{42}{7} \):
\[
\frac{x}{7} + 6 - 6 = \frac{47}{7} - \frac{42}{7}
\]
\[
\frac{x}{7} = \frac{5}{7}
\]

2. Multiply both sides by 7:
\[
7 \cdot \frac{x}{7} = 7 \cdot \frac{5}{7}
\]
\[
x = 5
\]

Solution: \( x = 5 \)

---

Equation 8: \( \frac{x}{7} - 7 = -\frac{41}{7} \)

1. Add 7 to both sides. Note that 7 can be written as \( \frac{49}{7} \):
\[
\frac{x}{7} - 7 + 7 = -\frac{41}{7} + \frac{49}{7}
\]
\[
\frac{x}{7} = \frac{8}{7}
\]

2. Multiply both sides by 7:
\[
7 \cdot \frac{x}{7} = 7 \cdot \frac{8}{7}
\]
\[
x = 8
\]

Solution: \( x = 8 \)

---

Equation 9: \( \frac{x}{8} - 8 = -7 \)

1. Add 8 to both sides:
\[
\frac{x}{8} - 8 + 8 = -7 + 8
\]
\[
\frac{x}{8} = 1
\]

2. Multiply both sides by 8:
\[
8 \cdot \frac{x}{8} = 8 \cdot 1
\]
\[
x = 8
\]

Solution: \( x = 8 \)

---

Equation 10: \( \frac{x}{3} + 5 = \frac{20}{3} \)

1. Subtract 5 from both sides. Note that 5 can be written as \( \frac{15}{3} \):
\[
\frac{x}{3} + 5 - 5 = \frac{20}{3} - \frac{15}{3}
\]
\[
\frac{x}{3} = \frac{5}{3}
\]

2. Multiply both sides by 3:
\[
3 \cdot \frac{x}{3} = 3 \cdot \frac{5}{3}
\]
\[
x = 5
\]

Solution: \( x = 5 \)

---

Equation 11: \( \frac{x}{3} - 7 = -\frac{17}{3} \)

1. Add 7 to both sides. Note that 7 can be written as \( \frac{21}{3} \):
\[
\frac{x}{3} - 7 + 7 = -\frac{17}{3} + \frac{21}{3}
\]
\[
\frac{x}{3} = \frac{4}{3}
\]

2. Multiply both sides by 3:
\[
3 \cdot \frac{x}{3} = 3 \cdot \frac{4}{3}
\]
\[
x = 4
\]

Solution: \( x = 4 \)

---

Equation 12: \( \frac{x}{3} + 3 = \frac{14}{3} \)

1. Subtract 3 from both sides. Note that 3 can be written as \( \frac{9}{3} \):
\[
\frac{x}{3} + 3 - 3 = \frac{14}{3} - \frac{9}{3}
\]
\[
\frac{x}{3} = \frac{5}{3}
\]

2. Multiply both sides by 3:
\[
3 \cdot \frac{x}{3} = 3 \cdot \frac{5}{3}
\]
\[
x = 5
\]

Solution: \( x = 5 \)

---

Equation 13: \( \frac{x}{8} - 5 = -\frac{19}{4} \)

1. Add 5 to both sides. Note that 5 can be written as \( \frac{40}{8} \) or \( \frac{20}{4} \):
\[
\frac{x}{8} - 5 + 5 = -\frac{19}{4} + \frac{20}{4}
\]
\[
\frac{x}{8} = \frac{1}{4}
\]

2. Eliminate the denominator by multiplying both sides by 8:
\[
8 \cdot \frac{x}{8} = 8 \cdot \frac{1}{4}
\]
\[
x = 2
\]

Solution: \( x = 2 \)

---

Equation 14: \( \frac{x}{6} - 6 = -\frac{16}{3} \)

1. Add 6 to both sides. Note that 6 can be written as \( \frac{18}{3} \):
\[
\frac{x}{6} - 6 + 6 = -\frac{16}{3} + \frac{18}{3}
\]
\[
\frac{x}{6} = \frac{2}{3}
\]

2. Eliminate the denominator by multiplying both sides by 6:
\[
6 \cdot \frac{x}{6} = 6 \cdot \frac{2}{3}
\]
\[
x = 4
\]

Solution: \( x = 4 \)

---

Equation 15: \( \frac{x}{8} - 6 = -\frac{21}{4} \)

1. Add 6 to both sides. Note that 6 can be written as \( \frac{24}{4} \):
\[
\frac{x}{8} - 6 + 6 = -\frac{21}{4} + \frac{24}{4}
\]
\[
\frac{x}{8} = \frac{3}{4}
\]

2. Eliminate the denominator by multiplying both sides by 8:
\[
8 \cdot \frac{x}{8} = 8 \cdot \frac{3}{4}
\]
\[
x = 6
\]

Solution: \( x = 6 \)

---

Final Answers:

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