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Let's go through the problem step by step.

### Problem Statement:
- **Whittier Drive** is needed to bring 80 students to a charity event.
- **Cars** and **vans** are the two types of vehicles available.
- **Cars** can carry 4 students each.
- **Vans** can carry 10 students each.

The equation to model the situation is:
\[ 4C + 10V = 80 \]
Where \( C \) is the number of cars and \( V \) is the number of vans.

### Task:
Select all statements that are true about the situation.

### Let's analyze the statements:

1. **If 12 cars go, then 2 vans are needed.**
   - Substituting \( C = 12 \) into the equation:
     \[
     4(12) + 10V = 80 \implies 48 + 10V = 80 \implies 10V = 32
     \]
     \[
     V = \frac{32}{10} = 3.2
     \]
   - \( V = 3.2 \) is not a whole number, so this statement is **False**.

2. **C = 1 and V = 4 are a pair of solutions to the equation.**
   - Substituting \( C = 1 \) and \( V = 4 \) into the equation:
     \[
     4(1) + 10(4) = 4 + 40 = 44
     \]
   - The sum is 44, not 80. Therefore, this statement is **False**.

3. **If 6 cars go and 11 vans go, there will be extra space.**
   - Substituting \( C = 6 \) and \( V = 11 \) into the equation:
     \[
     4(6) + 10(11) = 24 + 110 = 134
     \]
   - 134 students can be transported, which is more than 80, so there will be extra space. This statement is **True**.

4. **10 cars and 8 vans isn’t enough to transport all the students.**
   - Substituting \( C = 10 \) and \( V = 8 \) into the equation:
     \[
     4(10) + 10(8) = 40 + 80 = 120
     \]
   - 120 students can be transported, which is more than 80, so this statement is **False**.

5. **If 20 cars go, no vans are needed.**
   - Substituting \( C = 20 \) and \( V = 0 \) into the equation:
     \[
     4(20) + 10(0) = 80 + 0 = 80
     \]
   - 80 students can be transported with 20 cars and no vans, so this statement is **True**.

6. **8 vans and 8 cars are numbers that meet the constraints in this situation.**
   - Substituting \( C = 8 \) and \( V = 8 \) into the equation:
     \[
     4(8) + 10(8) = 32 + 80 = 112
     \]
   - 112 students can be transported, which is more than 80, so this statement is **False**.

### True Statements:
- **Statement 3** (If 6 cars go and 11 vans go, there will be extra space.)
- **Statement 5** (If 20 cars go, no vans are needed.)

Would you like more details or explanations?

### Follow-up Questions:
1. What would be the minimum number of vehicles required to transport all 80 students?
2. If 5 vans are used, how many cars are needed to meet the requirements?
3. Could the number of vehicles be optimized further?
4. What would happen if the capacity of each van increased to 12 students?
5. How would the scenario change if the number of students increased to 100?

### Tip:
Always verify the solutions by substituting the values back into the original equation to ensure the equation holds true.

Whittier Drive is needed to bring 80 students to a charity event. Cars and vans are the two types of vehicles available. Cars can carry 4 students each, and vans can carry 10 students each. The equation to model the situation is 4C + 10V = 80. Select all statements that are true about the situation.

Learn how to solve a linear equation to determine the number of cars and vans needed to transport 80 students to a charity event. This problem involves key algebraic concepts such as linear equations, substitution, and understanding solution constraints. Suitable for students in Grades 6-8.

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