Fraction Review: Addition, Subtraction, and Inequalities ... - Free Printable

Problem Overview:

The image contains a fraction review worksheet with three sections:
1. Addition and Subtraction of Fractions: Solve each problem by adding or subtracting fractions.
2. Comparison of Fraction Results: Compare the results of addition and subtraction problems using "<", ">", or "=".
3. Finding Missing Factors: Use inverse operations to find missing factors in equations involving fractions.

We will solve each section step by step.

---

Section 1: Addition and Subtraction of Fractions

#### Problem 1:
$$
\frac{1}{2} - \frac{1}{4} =
$$
- Step 1: Find a common denominator. The denominators are 2 and 4, so the least common denominator (LCD) is 4.
- Step 2: Rewrite the fractions with the common denominator:
$$
\frac{1}{2} = \frac{2}{4}, \quad \frac{1}{4} = \frac{1}{4}
$$
- Step 3: Subtract the numerators:
$$
\frac{2}{4} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4}
$$
- Answer:
$$
\frac{1}{2} - \frac{1}{4} = \frac{1}{4}
$$

#### Problem 2:
$$
\frac{4}{8} + \frac{4}{8} =
$$
- Step 1: The denominators are the same (8), so add the numerators directly:
$$
\frac{4}{8} + \frac{4}{8} = \frac{4 + 4}{8} = \frac{8}{8}
$$
- Step 2: Simplify the fraction:
$$
\frac{8}{8} = 1
$$
- Answer:
$$
\frac{4}{8} + \frac{4}{8} = 1
$$

#### Problem 3:
$$
\frac{3}{5} + \frac{4}{7} =
$$
- Step 1: Find the least common denominator (LCD). The denominators are 5 and 7, so the LCD is \(5 \times 7 = 35\).
- Step 2: Rewrite the fractions with the common denominator:
$$
\frac{3}{5} = \frac{3 \times 7}{5 \times 7} = \frac{21}{35}, \quad \frac{4}{7} = \frac{4 \times 5}{7 \times 5} = \frac{20}{35}
$$
- Step 3: Add the numerators:
$$
\frac{21}{35} + \frac{20}{35} = \frac{21 + 20}{35} = \frac{41}{35}
$$
- Answer:
$$
\frac{3}{5} + \frac{4}{7} = \frac{41}{35}
$$

#### Problem 4:
$$
\frac{3}{5} - \frac{1}{3} =
$$
- Step 1: Find the least common denominator (LCD). The denominators are 5 and 3, so the LCD is \(5 \times 3 = 15\).
- Step 2: Rewrite the fractions with the common denominator:
$$
\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}, \quad \frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}
$$
- Step 3: Subtract the numerators:
$$
\frac{9}{15} - \frac{5}{15} = \frac{9 - 5}{15} = \frac{4}{15}
$$
- Answer:
$$
\frac{3}{5} - \frac{1}{3} = \frac{4}{15}
$$

#### Problem 5:
$$
\frac{3}{7} - \frac{1}{4} =
$$
- Step 1: Find the least common denominator (LCD). The denominators are 7 and 4, so the LCD is \(7 \times 4 = 28\).
- Step 2: Rewrite the fractions with the common denominator:
$$
\frac{3}{7} = \frac{3 \times 4}{7 \times 4} = \frac{12}{28}, \quad \frac{1}{4} = \frac{1 \times 7}{4 \times 7} = \frac{7}{28}
$$
- Step 3: Subtract the numerators:
$$
\frac{12}{28} - \frac{7}{28} = \frac{12 - 7}{28} = \frac{5}{28}
$$
- Answer:
$$
\frac{3}{7} - \frac{1}{4} = \frac{5}{28}
$$

---

Section 2: Comparison of Fraction Results

#### Problem 6:
$$
\frac{6}{4} - \frac{3}{20} \quad \text{and} \quad \frac{4}{3} - \frac{3}{20}
$$
- Step 1: Simplify each expression separately.
- For \(\frac{6}{4} - \frac{3}{20}\):
- Simplify \(\frac{6}{4}\): \(\frac{6}{4} = \frac{3}{2}\).
- Find the LCD of 2 and 20, which is 20.
- Rewrite the fractions:
$$
\frac{3}{2} = \frac{3 \times 10}{2 \times 10} = \frac{30}{20}, \quad \frac{3}{20} = \frac{3}{20}
$$
- Subtract:
$$
\frac{30}{20} - \frac{3}{20} = \frac{30 - 3}{20} = \frac{27}{20}
$$
- For \(\frac{4}{3} - \frac{3}{20}\):
- Find the LCD of 3 and 20, which is 60.
- Rewrite the fractions:
$$
\frac{4}{3} = \frac{4 \times 20}{3 \times 20} = \frac{80}{60}, \quad \frac{3}{20} = \frac{3 \times 3}{20 \times 3} = \frac{9}{60}
$$
- Subtract:
$$
\frac{80}{60} - \frac{9}{60} = \frac{80 - 9}{60} = \frac{71}{60}
$$
- Step 2: Compare \(\frac{27}{20}\) and \(\frac{71}{60}\).
- Convert both fractions to have a common denominator (LCD = 60):
$$
\frac{27}{20} = \frac{27 \times 3}{20 \times 3} = \frac{81}{60}, \quad \frac{71}{60} = \frac{71}{60}
$$
- Compare the numerators:
$$
81 > 71 \implies \frac{27}{20} > \frac{71}{60}
$$
- Answer:
$$
\frac{6}{4} - \frac{3}{20} > \frac{4}{3} - \frac{3}{20}
$$

#### Problem 7:
$$
\frac{6}{10} + \frac{1}{4} \quad \text{and} \quad \frac{2}{4} + \frac{7}{12}
$$
- Step 1: Simplify each expression separately.
- For \(\frac{6}{10} + \frac{1}{4}\):
- Simplify \(\frac{6}{10}\): \(\frac{6}{10} = \frac{3}{5}\).
- Find the LCD of 5 and 4, which is 20.
- Rewrite the fractions:
$$
\frac{3}{5} = \frac{3 \times 4}{5 \times 4} = \frac{12}{20}, \quad \frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}
$$
- Add:
$$
\frac{12}{20} + \frac{5}{20} = \frac{12 + 5}{20} = \frac{17}{20}
$$
- For \(\frac{2}{4} + \frac{7}{12}\):
- Simplify \(\frac{2}{4}\): \(\frac{2}{4} = \frac{1}{2}\).
- Find the LCD of 2 and 12, which is 12.
- Rewrite the fractions:
$$
\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}, \quad \frac{7}{12} = \frac{7}{12}
$$
- Add:
$$
\frac{6}{12} + \frac{7}{12} = \frac{6 + 7}{12} = \frac{13}{12}
$$
- Step 2: Compare \(\frac{17}{20}\) and \(\frac{13}{12}\).
- Convert both fractions to have a common denominator (LCD = 60):
$$
\frac{17}{20} = \frac{17 \times 3}{20 \times 3} = \frac{51}{60}, \quad \frac{13}{12} = \frac{13 \times 5}{12 \times 5} = \frac{65}{60}
$$
- Compare the numerators:
$$
51 < 65 \implies \frac{17}{20} < \frac{13}{12}
$$
- Answer:
$$
\frac{6}{10} + \frac{1}{4} < \frac{2}{4} + \frac{7}{12}
$$

#### Problem 8:
$$
3 \frac{3}{4} - 3 \frac{5}{7} \quad \text{and} \quad 6 \frac{7}{8} - 3 \frac{2}{4}
$$
- Step 1: Convert mixed numbers to improper fractions.
- For \(3 \frac{3}{4}\):
$$
3 \frac{3}{4} = \frac{3 \times 4 + 3}{4} = \frac{15}{4}
$$
- For \(3 \frac{5}{7}\):
$$
3 \frac{5}{7} = \frac{3 \times 7 + 5}{7} = \frac{26}{7}
$$
- For \(6 \frac{7}{8}\):
$$
6 \frac{7}{8} = \frac{6 \times 8 + 7}{8} = \frac{55}{8}
$$
- For \(3 \frac{2}{4}\):
$$
3 \frac{2}{4} = \frac{3 \times 4 + 2}{4} = \frac{14}{4}
$$
- Step 2: Perform the subtractions.
- For \(\frac{15}{4} - \frac{26}{7}\):
- Find the LCD of 4 and 7, which is 28.
- Rewrite the fractions:
$$
\frac{15}{4} = \frac{15 \times 7}{4 \times 7} = \frac{105}{28}, \quad \frac{26}{7} = \frac{26 \times 4}{7 \times 4} = \frac{104}{28}
$$
- Subtract:
$$
\frac{105}{28} - \frac{104}{28} = \frac{105 - 104}{28} = \frac{1}{28}
$$
- For \(\frac{55}{8} - \frac{14}{4}\):
- Simplify \(\frac{14}{4}\): \(\frac{14}{4} = \frac{7}{2}\).
- Find the LCD of 8 and 2, which is 8.
- Rewrite the fractions:
$$
\frac{55}{8} = \frac{55}{8}, \quad \frac{7}{2} = \frac{7 \times 4}{2 \times 4} = \frac{28}{8}
$$
- Subtract:
$$
\frac{55}{8} - \frac{28}{8} = \frac{55 - 28}{8} = \frac{27}{8}
$$
- Step 3: Compare \(\frac{1}{28}\) and \(\frac{27}{8}\).
- Clearly, \(\frac{1}{28}\) is much smaller than \(\frac{27}{8}\).
- Answer:
$$
3 \frac{3}{4} - 3 \frac{5}{7} < 6 \frac{7}{8} - 3 \frac{2}{4}
$$

#### Problem 9:
$$
5 \frac{9}{6} + 5 \frac{2}{3} \quad \text{and} \quad 8 \frac{7}{9} - 4 \frac{1}{3}
$$
- Step 1: Convert mixed numbers to improper fractions.
- For \(5 \frac{9}{6}\):
$$
5 \frac{9}{6} = \frac{5 \times 6 + 9}{6} = \frac{39}{6}
$$
- For \(5 \frac{2}{3}\):
$$
5 \frac{2}{3} = \frac{5 \times 3 + 2}{3} = \frac{17}{3}
$$
- For \(8 \frac{7}{9}\):
$$
8 \frac{7}{9} = \frac{8 \times 9 + 7}{9} = \frac{79}{9}
$$
- For \(4 \frac{1}{3}\):
$$
4 \frac{1}{3} = \frac{4 \times 3 + 1}{3} = \frac{13}{3}
$$
- Step 2: Perform the operations.
- For \(\frac{39}{6} + \frac{17}{3}\):
- Simplify \(\frac{39}{6}\): \(\frac{39}{6} = \frac{13}{2}\).
- Find the LCD of 2 and 3, which is 6.
- Rewrite the fractions:
$$
\frac{13}{2} = \frac{13 \times 3}{2 \times 3} = \frac{39}{6}, \quad \frac{17}{3} = \frac{17 \times 2}{3 \times 2} = \frac{34}{6}
$$
- Add:
$$
\frac{39}{6} + \frac{34}{6} = \frac{39 + 34}{6} = \frac{73}{6}
$$
- For \(\frac{79}{9} - \frac{13}{3}\):
- Find the LCD of 9 and 3, which is 9.
- Rewrite the fractions:
$$
\frac{79}{9} = \frac{79}{9}, \quad \frac{13}{3} = \frac{13 \times 3}{3 \times 3} = \frac{39}{9}
$$
- Subtract:
$$
\frac{79}{9} - \frac{39}{9} = \frac{79 - 39}{9} = \frac{40}{9}
$$
- Step 3: Compare \(\frac{73}{6}\) and \(\frac{40}{9}\).
- Convert both fractions to have a common denominator (LCD = 54):
$$
\frac{73}{6} = \frac{73 \times 9}{6 \times 9} = \frac{657}{54}, \quad \frac{40}{9} = \frac{40 \times 6}{9 \times 6} = \frac{240}{54}
$$
- Compare the numerators:
$$
657 > 240 \implies \frac{73}{6} > \frac{40}{9}
$$
- Answer:
$$
5 \frac{9}{6} + 5 \frac{2}{3} > 8 \frac{7}{9} - 4 \frac{1}{3}
$$

#### Problem 10:
$$
4 \frac{1}{2} - 1 \frac{1}{8} \quad \text{and} \quad 3 \frac{2}{3} + 5 \frac{3}{6}
$$
- Step 1: Convert mixed numbers to improper fractions.
- For \(4 \frac{1}{2}\):
$$
4 \frac{1}{2} = \frac{4 \times 2 + 1}{2} = \frac{9}{2}
$$
- For \(1 \frac{1}{8}\):
$$
1 \frac{1}{8} = \frac{1 \times 8 + 1}{8} = \frac{9}{8}
$$
- For \(3 \frac{2}{3}\):
$$
3 \frac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{11}{3}
$$
- For \(5 \frac{3}{6}\):
- Simplify \(\frac{3}{6}\): \(\frac{3}{6} = \frac{1}{2}\).
- Convert \(5 \frac{1}{2}\) to an improper fraction:
$$
5 \frac{1}{2} = \frac{5 \times 2 + 1}{2} = \frac{11}{2}
$$
- Step 2: Perform the operations.
- For \(\frac{9}{2} - \frac{9}{8}\):
- Find the LCD of 2 and 8, which is 8.
- Rewrite the fractions:
$$
\frac{9}{2} = \frac{9 \times 4}{2 \times 4} = \frac{36}{8}, \quad \frac{9}{8} = \frac{9}{8}
$$
- Subtract:
$$
\frac{36}{8} - \frac{9}{8} = \frac{36 - 9}{8} = \frac{27}{8}
$$
- For \(\frac{11}{3} + \frac{11}{2}\):
- Find the LCD of 3 and 2, which is 6.
- Rewrite the fractions:
$$
\frac{11}{3} = \frac{11 \times 2}{3 \times 2} = \frac{22}{6}, \quad \frac{11}{2} = \frac{11 \times 3}{2 \times 3} = \frac{33}{6}
$$
- Add:
$$
\frac{22}{6} + \frac{33}{6} = \frac{22 + 33}{6} = \frac{55}{6}
$$
- Step 3: Compare \(\frac{27}{8}\) and \(\frac{55}{6}\).
- Convert both fractions to have a common denominator (LCD = 24):
$$
\frac{27}{8} = \frac{27 \times 3}{8 \times 3} = \frac{81}{24}, \quad \frac{55}{6} = \frac{55 \times 4}{6 \times 4} = \frac{220}{24}
$$
- Compare the numerators:
$$
81 < 220 \implies \frac{27}{8} < \frac{55}{6}
$$
- Answer:
$$
4 \frac{1}{2} - 1 \frac{1}{8} < 3 \frac{2}{3} + 5 \frac{3}{6}
$$

---

Section 3: Finding Missing Factors

#### Problem 1:
$$
1 \frac{1}{2} - \boxed{\phantom{0}} = 2 \frac{7}{8}
$$
- Step 1: Convert mixed numbers to improper fractions.
- For \(1 \frac{1}{2}\):
$$
1 \frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}
$$
- For \(2 \frac{7}{8}\):
$$
2 \frac{7}{8} = \frac{2 \times 8 + 7}{8} = \frac{23}{8}
$$
- Step 2: Let the missing factor be \(x\). The equation becomes:
$$
\frac{3}{2} - x = \frac{23}{8}
$$
- Step 3: Solve for \(x\):
- Isolate \(x\):
$$
x = \frac{3}{2} - \frac{23}{8}
$$
- Find the LCD of 2 and 8, which is 8.
- Rewrite the fractions:
$$
\frac{3}{2} = \frac{3 \times 4}{2 \times 4} = \frac{12}{8}, \quad \frac{23}{8} = \frac{23}{8}
$$
- Subtract:
$$
x = \frac{12}{8} - \frac{23}{8} = \frac{12 - 23}{8} = \frac{-11}{8}
$$
- Answer:
$$
\boxed{\frac{-11}{8}}
$$

#### Problem 2:
$$
\boxed{\phantom{0}} + 1 \frac{1}{2} = 11
$$
- Step 1: Convert the mixed number to an improper fraction.
- For \(1 \frac{1}{2}\):
$$
1 \frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}
$$
- Step 2: Let the missing factor be \(x\). The equation becomes:
$$
x + \frac{3}{2} = 11
$$
- Step 3: Solve for \(x\):
- Isolate \(x\):
$$
x = 11 - \frac{3}{2}
$$
- Convert 11 to a fraction with denominator 2:
$$
11 = \frac{11 \times 2}{1 \times 2} = \frac{22}{2}
$$
- Subtract:
$$
x = \frac{22}{2} - \frac{3}{2} = \frac{22 - 3}{2} = \frac{19}{2}
$$
- Answer:
$$
\boxed{\frac{19}{2}}
$$

#### Problem 3:
$$
\boxed{\phantom{0}} + \frac{7}{8} = 13 \frac{3}{8}
$$
- Step 1: Convert the mixed number to an improper fraction.
- For \(13 \frac{3}{8}\):
$$
13 \frac{3}{8} = \frac{13 \times 8 + 3}{8} = \frac{107}{8}
$$
- Step 2: Let the missing factor be \(x\). The equation becomes:
$$
x + \frac{7}{8} = \frac{107}{8}
$$
- Step 3: Solve for \(x\):
- Isolate \(x\):
$$
x = \frac{107}{8} - \frac{7}{8}
$$
- Subtract:
$$
x = \frac{107 - 7}{8} = \frac{100}{8}
$$
- Simplify:
$$
\frac{100}{8} = \frac{25}{2}
$$
- Answer:
$$
\boxed{\frac{25}{2}}
$$

#### Problem 4:
$$
7 \frac{5}{8} - \boxed{\phantom{0}} = 5 \frac{3}{8}
$$
- Step 1: Convert the mixed numbers to improper fractions.
- For \(7 \frac{5}{8}\):
$$
7 \frac{5}{8} = \frac{7 \times 8 + 5}{8} = \frac{61}{8}
$$
- For \(5 \frac{3}{8}\):
$$
5 \frac{3}{8} = \frac{5 \times 8 + 3}{8} = \frac{43}{8}
$$
- Step 2: Let the missing factor be \(x\). The equation becomes:
$$
\frac{61}{8} - x = \frac{43}{8}
$$
- Step 3: Solve for \(x\):
- Isolate \(x\):
$$
x = \frac{61}{8} - \frac{43}{8}
$$
- Subtract:
$$
x = \frac{61 - 43}{8} = \frac{18}{8}
$$
- Simplify:
$$
\frac{18}{8} = \frac{9}{4}
$$
- Answer:
$$
\boxed{\frac{9}{4}}
$$

---

Final Answers:

1. \(\frac{1}{2} - \frac{1}{4} = \frac{1}{4}\)
2. \(\frac{4}{8} + \frac{4}{8} = 1\)
3. \(\frac{3}{5} + \frac{4}{7} = \frac{41}{35}\)
4. \(\frac{3}{5} - \frac{1}{3} = \frac{4}{15}\)
5. \(\frac{3}{7} - \frac{1}{4} = \frac{5}{28}\)
6. \(\frac{6}{4} - \frac{3}{20} > \frac{4}{3} - \frac{3}{20}\)
7. \(\frac{6}{10} + \frac{1}{4} < \frac{2}{4} + \frac{7}{12}\)
8. \(3 \frac{3}{4} - 3 \frac{5}{7} < 6 \frac{7}{8} - 3 \frac{2}{4}\)
9. \(5 \frac{9}{6} + 5 \frac{2}{3} > 8 \frac{7}{9} - 4 \frac{1}{3}\)
10. \(4 \frac{1}{2} - 1 \frac{1}{8} < 3 \frac{2}{3} + 5 \frac{3}{6}\)
11. \(1 \frac{1}{2} - \boxed{\frac{-11}{8}} = 2 \frac{7}{8}\)
12. \(\boxed{\frac{19}{2}} + 1 \frac{1}{2} = 11\)
13. \(\boxed{\frac{25}{2}} + \frac{7}{8} = 13 \frac{3}{8}\)
14. \(7 \frac{5}{8} - \boxed{\frac{9}{4}} = 5 \frac{3}{8}\)

$$
\boxed{\text{All problems solved.}}
$$