To add the fractions \( \frac{1}{2}(x + 4) + \frac{1}{4}(4x - 2) \), we first distribute each term.
- Distribute \( \frac{1}{2} \) into \( (x + 4) \):
\[ \frac{1}{2}(x + 4) = \frac{1}{2}x + \frac{1}{2} \cdot 4 = \frac{1}{2}x + 2 \]
- Distribute \( \frac{1}{4} \) into \( (4x - 2) \):
\[ \frac{1}{4}(4x - 2) = \frac{1}{4} \cdot 4x - \frac{1}{4} \cdot 2 = x - \frac{1}{2} \]
- Now we combine the two results:
\[ \left( \frac{1}{2}x + 2 \right) + \left( x - \frac{1}{2} \right) \]
Combine like terms:
- For \( x \) terms: \( \frac{1}{2}x + x = \frac{1}{2}x + \frac{2}{2}x = \frac{3}{2}x \)
- For constant terms: \( 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \)
Putting it all together, we have:
\[ \frac{3}{2}x + \frac{3}{2} \]
Now, let's express the constant term \( \frac{3}{2} \) in mixed number form.
Since \( \frac{3}{2} = 1 \frac{1}{2} \), we have:
The final answer is:
\[ \frac{3}{2}x + 1 \frac{1}{2} \]