The system has a unique solution for any given b1 and b2
The system will have infinitely many solutions for any given b1 and b2
Whether or not a solution exists depends on the given b1 and b2
The system would have no solution for any values of b1 and b2
Answer is Right!
Answer is Wrong!
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.