To solve the equation \(2a + b = 430\) for \(a\) and \(b\), we need additional information or a second equation. With one equation and two variables, there are infinitely many solutions. However, we can express one variable in terms of the other.

### Solving for \(b\):
\[ b = 430 - 2a \]

This equation gives us the value of \(b\) for any value of \(a\).

### Solving for \(a\):
\[ a = \frac{430 - b}{2} \]

This equation gives us the value of \(a\) for any value of \(b\).

Would you like to provide more information or have any specific values for \(a\) or \(b\) in mind? 

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Here are five related questions:
1. What happens if you add a second equation to this system?
2. How do you graph the equation \(2a + b = 430\)?
3. Can you solve this system if \(a\) or \(b\) is set to a specific value?
4. What are the possible integer values for \(a\) and \(b\) if both must be positive?
5. How would the solution change if the equation was \(2a + b = c\) for a different constant \(c\)?

**Tip:** In linear equations with two variables, finding a solution set often involves either substituting known values or adding another equation to create a solvable system.

Learn how to solve the linear equation 2a + b = 430 by expressing one variable in terms of the other. This problem is ideal for students in Grades 6-8 and covers key algebraic concepts such as linear equations and variable isolation.

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