Fraction addition and subtraction practice worksheet for students.

Let’s solve each problem step by step. All these problems involve adding or subtracting fractions that have the same denominator (bottom number). When denominators are the same, you just add or subtract the numerators (top numbers) and keep the denominator the same.

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1.
$\frac{1}{3} + \frac{2}{3}$
Add top numbers: $1 + 2 = 3$
Keep bottom number: $3$
→ $\frac{3}{3} = 1$

2.
$\frac{1}{4} + \frac{3}{4}$
$1 + 3 = 4$ → $\frac{4}{4} = 1$

3.
$\frac{3}{5} + \frac{2}{5}$
$3 + 2 = 5$ → $\frac{5}{5} = 1$

4.
$\frac{3}{7} + \frac{5}{7}$
$3 + 5 = 8$ → $\frac{8}{7}$
This is an improper fraction — it’s okay to leave it like this unless told to simplify or convert to mixed number. But since no instruction, we’ll leave as $\frac{8}{7}$

Wait — let me double-check: 3+5=8, yes. Denominator stays 7. So $\frac{8}{7}$ is correct.

But sometimes teachers want simplified form. $\frac{8}{7} = 1\frac{1}{7}$, but again — unless asked, we can leave as improper. Let’s check other problems for consistency.

Looking at problem 7: $\frac{2}{9} + \frac{7}{9} = \frac{9}{9} = 1$, so they expect simplification when possible.

So for #4: $\frac{8}{7}$ is already in simplest form? Yes, because 8 and 7 share no common factors. So we can leave it as $\frac{8}{7}$, or write as $1\frac{1}{7}$. Since the worksheet doesn’t specify, I’ll go with simplified fraction or whole number if possible. For now, I’ll write as $\frac{8}{7}$, but note that some might prefer mixed number. Let’s see what others do.

Actually, looking at all answers, most will be whole numbers or simple fractions. Let’s proceed and adjust if needed.

5.
$\frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$ (simplify by dividing numerator and denominator by 4)

6.
$\frac{1}{2} + \frac{2}{2} = \frac{3}{2}$ → which is $1\frac{1}{2}$, but again, maybe leave as $\frac{3}{2}$? Wait — $\frac{2}{2} = 1$, so $\frac{1}{2} + 1 = 1\frac{1}{2}$. But since it's written as fractions, perhaps keep as $\frac{3}{2}$. However, let’s check problem 13: $\frac{3}{5} - \frac{2}{5} = \frac{1}{5}$, so they don’t always reduce to mixed numbers. I think for consistency, we’ll reduce fractions where possible, but not necessarily convert to mixed numbers unless it’s a whole number.

Actually, let’s look at problem 7: $\frac{2}{9} + \frac{7}{9} = \frac{9}{9} = 1$, so they do simplify to whole numbers when possible.

For #6: $\frac{3}{2}$ is fine, or $1\frac{1}{2}$. I think for elementary level, they might prefer mixed numbers. But since the worksheet doesn’t specify, and to be safe, I’ll write reduced fractions, and if it’s greater than 1, I’ll write as mixed number only if it’s obvious. Actually, let’s follow standard practice: reduce fractions, and if improper, convert to mixed number.

I recall that in many worksheets like this, they expect simplified answers, including converting improper fractions to mixed numbers.

Let me redo with that in mind.

Revised plan:
- Add/subtract numerators.
- Keep denominator.
- Simplify fraction if possible (divide numerator and denominator by GCF).
- If result is improper fraction (numerator > denominator), convert to mixed number.

Let’s apply that.

1. $\frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1$

2. $\frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$

3. $\frac{3}{5} + \frac{2}{5} = \frac{5}{5} = 1$

4. $\frac{3}{7} + \frac{5}{7} = \frac{8}{7} = 1\frac{1}{7}$

5. $\frac{3}{8} + \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$

6. $\frac{1}{2} + \frac{2}{2} = \frac{3}{2} = 1\frac{1}{2}$

7. $\frac{2}{9} + \frac{7}{9} = \frac{9}{9} = 1$

Now subtraction side:

8. $\frac{4}{7} - \frac{2}{7} = \frac{2}{7}$

9. $\frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ (simplify by dividing by 2)

10. $\frac{5}{8} - \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$

11. $\frac{5}{9} - \frac{1}{9} = \frac{4}{9}$

12. $\frac{6}{7} - \frac{2}{7} = \frac{4}{7}$

13. $\frac{3}{5} - \frac{2}{5} = \frac{1}{5}$

14. $\frac{2}{6} - \frac{1}{6} = \frac{1}{6}$

Wait, problem 14: $\frac{2}{6} - \frac{1}{6} = \frac{1}{6}$, but $\frac{2}{6}$ can be simplified to $\frac{1}{3}$, but since we’re subtracting, better to keep same denominator. After subtraction, $\frac{1}{6}$ is already simplified.

Also, problem 9: $\frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$, yes.

Now, let me list all answers clearly:

1. 1
2. 1
3. 1
4. $1\frac{1}{7}$
5. $\frac{1}{2}$
6. $1\frac{1}{2}$
7. 1
8. $\frac{2}{7}$
9. $\frac{1}{3}$
10. $\frac{1}{2}$
11. $\frac{4}{9}$
12. $\frac{4}{7}$
13. $\frac{1}{5}$
14. $\frac{1}{6}$

I think that’s correct. Let me verify a couple more.

Problem 4: 3/7 + 5/7 = 8/7 = 1 and 1/7 → yes.

Problem 6: 1/2 + 2/2 = 1/2 + 1 = 1.5 = 3/2 = 1 1/2 → yes.

Problem 9: 3/6 is 1/2, minus 1/6 is 3/6 - 1/6 = 2/6 = 1/3 → yes.

All good.

Final Answer:
1. 1
2. 1
3. 1
4. $1\frac{1}{7}$
5. $\frac{1}{2}$
6. $1\frac{1}{2}$
7. 1
8. $\frac{2}{7}$
9. $\frac{1}{3}$
10. $\frac{1}{2}$
11. $\frac{4}{9}$
12. $\frac{4}{7}$
13. $\frac{1}{5}$
14. $\frac{1}{6}$