Sure, let's walk through each of these addition and subtraction problems one-by-one.
### 1. Addition Sums
a. [tex]\(\frac{3}{4} + \frac{1}{4}\)[/tex]
To add these fractions, we add the numerators since the denominators are the same:
[tex]\[ \frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1 \][/tex]
b. [tex]\(\frac{2}{5} + \frac{1}{5}\)[/tex]
Similarly, add the numerators because the denominators are the same:
[tex]\[ \frac{2}{5} + \frac{1}{5} = \frac{2+1}{5} = \frac{3}{5} \][/tex]
c. [tex]\(\frac{4}{7} + \frac{1}{7}\)[/tex]
Add the numerators:
[tex]\[ \frac{4}{7} + \frac{1}{7} = \frac{4+1}{7} = \frac{5}{7} \][/tex]
d. [tex]\(2 \frac{3}{8} + \frac{3}{8}\)[/tex]
First, convert the mixed number into an improper fraction:
[tex]\[ 2 \frac{3}{8} = \frac{2 \times 8 + 3}{8} = \frac{16 + 3}{8} = \frac{19}{8} \][/tex]
Now add:
[tex]\[ \frac{19}{8} + \frac{3}{8} = \frac{19+3}{8} = \frac{22}{8} = \frac{11}{4} \][/tex]
e. [tex]\(1 \frac{2}{4} + \frac{1}{4}\)[/tex]
Convert to an improper fraction:
[tex]\[ 1 \frac{2}{4} = \frac{1 \times 4 + 2}{4} = \frac{4 + 2}{4} = \frac{6}{4} \][/tex]
Now add:
[tex]\[ \frac{6}{4} + \frac{1}{4} = \frac{6+1}{4} = \frac{7}{4} \][/tex]
f. [tex]\(4 \frac{1}{8} + \frac{4}{8}\)[/tex]
Convert to an improper fraction:
[tex]\[ 4 \frac{1}{8} = \frac{4 \times 8 + 1}{8} = \frac{32 + 1}{8} = \frac{33}{8} \][/tex]
Now add:
[tex]\[ \frac{33}{8} + \frac{4}{8} = \frac{33+4}{8} = \frac{37}{8} \][/tex]
### 2. Subtraction Sums
a. [tex]\(\frac{3}{4} - \frac{1}{4}\)[/tex]
Subtract the numerators:
[tex]\[ \frac{3}{4} - \frac{1}{4} = \frac{3-1}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
b. [tex]\(\frac{2}{5} - \frac{1}{5}\)[/tex]
Subtract the numerators:
[tex]\[ \frac{2}{5} - \frac{1}{5} = \frac{2-1}{5} = \frac{1}{5} \][/tex]
c. [tex]\(\frac{4}{7} - \frac{1}{7}\)[/tex]
Subtract the numerators:
[tex]\[ \frac{4}{7} - \frac{1}{7} = \frac{4-1}{7} = \frac{3}{7} \][/tex]
