https://r2.erweima.ai/mathbot/image/17264096736210nbbrfd5.jpg

The problem in the image is stated in French. Here is a translation and breakdown of the problem to solve:

### Problem Translation:
- A group of friends spent €390 at a restaurant.
- Some (group A) spent €30 each.
- The others (group B) spent €20 each.
- How many people could there be in groups A and B?

**Note:** The problem states that **7 different answers** are possible.

### Setting Up the Equations:
Let:
- \( x \) = number of people in group A (each spends €30)
- \( y \) = number of people in group B (each spends €20)

The total spent by both groups is:
\[ 30x + 20y = 390 \]

### Simplifying the Equation:
Divide the entire equation by 10:
\[ 3x + 2y = 39 \]

### Finding Integer Solutions:
We need integer solutions for \( x \) and \( y \) where both \( x \) and \( y \) are non-negative:
\[ y = \frac{39 - 3x}{2} \]

For \( y \) to be an integer, \( 39 - 3x \) must be even. This implies that \( 3x \) must have the same parity (odd/even nature) as 39, which means \( x \) must be odd.

### Possible Values for \( x \) and \( y \):
Let's find values of \( x \) (odd numbers) that give integer \( y \):

1. **\( x = 1 \):**
   \[ y = \frac{39 - 3 \times 1}{2} = \frac{39 - 3}{2} = \frac{36}{2} = 18 \]
   Solution: \( x = 1, y = 18 \)

2. **\( x = 3 \):**
   \[ y = \frac{39 - 3 \times 3}{2} = \frac{39 - 9}{2} = \frac{30}{2} = 15 \]
   Solution: \( x = 3, y = 15 \)

3. **\( x = 5 \):**
   \[ y = \frac{39 - 3 \times 5}{2} = \frac{39 - 15}{2} = \frac{24}{2} = 12 \]
   Solution: \( x = 5, y = 12 \)

4. **\( x = 7 \):**
   \[ y = \frac{39 - 3 \times 7}{2} = \frac{39 - 21}{2} = \frac{18}{2} = 9 \]
   Solution: \( x = 7, y = 9 \)

5. **\( x = 9 \):**
   \[ y = \frac{39 - 3 \times 9}{2} = \frac{39 - 27}{2} = \frac{12}{2} = 6 \]
   Solution: \( x = 9, y = 6 \)

6. **\( x = 11 \):**
   \[ y = \frac{39 - 3 \times 11}{2} = \frac{39 - 33}{2} = \frac{6}{2} = 3 \]
   Solution: \( x = 11, y = 3 \)

7. **\( x = 13 \):**
   \[ y = \frac{39 - 3 \times 13}{2} = \frac{39 - 39}{2} = \frac{0}{2} = 0 \]
   Solution: \( x = 13, y = 0 \)

So, the 7 different possible solutions are:
1. \( (x = 1, y = 18) \)
2. \( (x = 3, y = 15) \)
3. \( (x = 5, y = 12) \)
4. \( (x = 7, y = 9) \)
5. \( (x = 9, y = 6) \)
6. \( (x = 11, y = 3) \)
7. \( (x = 13, y = 0) \)

Would you like a detailed explanation of any of these steps?

### Relative Questions:
1. How would the equation change if the amounts spent by groups A and B were different?
2. What other methods could be used to find integer solutions to this problem?
3. Can there be more than 7 solutions if different constraints are applied?
4. How does changing the total amount spent affect the number of solutions?
5. What are the practical implications of solving such a problem in real-life situations?

### Tip:
When dealing with integer solutions in equations, look for ways to simplify the equation to identify the relationships between variables easily.

A group of friends spent €390 at a restaurant. Some (group A) spent €30 each. The others (group B) spent €20 each. How many people could there be in groups A and B? There are 7 possible solutions.

Learn how to solve a linear equation where a group of friends spent €390 at a restaurant. Group A spent €30 each and Group B spent €20 each. This problem explores different integer solutions with 7 possible answers, covering linear equations and integer solutions. Suitable for Grades 7-9.

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