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Let's break down the problem and create expressions for both Libby and Jim's sweets after the changes:

### Initial conditions:
- **Libby** starts with 20 sweets.
- **Jim** also starts with 20 sweets.

### Actions:
1. **Libby gives Jim _x_ sweets.**
   - After giving _x_ sweets, Libby will have \( 20 - x \) sweets.
   - Jim will have \( 20 + x \) sweets.

2. **Libby eats 3 of her sweets.**
   - After eating 3 sweets, Libby will have \( (20 - x) - 3 = 17 - x \) sweets.

3. **Jim eats half of his sweets.**
   - Jim's sweets after eating half will be \( \frac{1}{2}(20 + x) = \frac{20 + x}{2} \).

### Final expressions:
- **Libby's final number of sweets:** \( 17 - x \)
- **Jim's final number of sweets:** \( \frac{20 + x}{2} \)

Would you like more details on any specific part of this solution? 

Here are some related questions for further exploration:
1. What happens if Libby gives Jim all her sweets?
2. If \( x = 5 \), how many sweets will each person have in the end?
3. How would the expressions change if Jim gave sweets to Libby instead?
4. What if Libby ate twice the number of sweets she gave to Jim?
5. How would the result change if Jim ate one-third of his sweets instead of half?

**Tip:** Always double-check the order of operations in algebraic expressions when forming them from word problems!

Libby has 20 sweets. Jim also has 20 sweets. Libby gives Jim x sweets. Libby then eats 3 of her sweets. Jim then eats half of his sweets. Write expressions for the number of sweets Libby and Jim now have.

Learn how to write algebraic expressions to calculate the number of sweets Libby and Jim have after sharing and consuming. This problem covers basic algebraic concepts, including expressions with variables. Suitable for students in Grades 6-8.

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