The correct answer is: The system will have infinitely many solutions for any given b1 and b2.
To solve a system of equations, we can use the elimination method. In this method, we eliminate one of the variables by adding or subtracting the equations in such a way that the terms with that variable cancel out. In this case, we can eliminate $x$ by adding the equations together. This gives us the equation $6y + 5z = b1 + b2$. We can then solve for $y$ and $z$ in terms of $b1$ and $b2$. This gives us the solutions $y = \frac{b1 + b2 – 5z}{6}$ and $z = \frac{b1 – b2}{6}$. As you can see, there are infinitely many solutions, since $y$ and $z$ can take on any value for any given $b1$ and $b2$.
Here is a brief explanation of each option:
- Option A: The system has a unique solution for any given b1 and b2. This is not true, as we have shown that there are infinitely many solutions.
- Option B: The system will have infinitely many solutions for any given b1 and b2. This is true, as we have shown that there are infinitely many solutions.
- Option C: Whether or not a solution exists depends on the given b1 and b2. This is not true, as we have shown that there are always solutions, regardless of the values of $b1$ and $b2$.
- Option D: The system would have no solution for any values of b1 and b2. This is not true, as we have shown that there are always solutions.