For what value of `alpha,` system of equations `alpha x + 4y = alpha - 3, 12 x + alpha y = alpha ` will have a unique solution ?

Options

A. `alpha ne pm 6`
B. `alpha ne pm 5`
C. `alpha = 6`
D. None of these

Answer

To find the value of `alpha` for which the system of equations will have a unique solution, we need to analyze the given equations:

1. \( \alpha x + 4y = \alpha - 3 \) (Equation 1)
2. \( 12x + \alpha y = \alpha \) (Equation 2)

### Step 1: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a = \alpha \)
- \( b = 4 \)
- For Equation 2:
- \( l = 12 \)
- \( m = \alpha \)

### Step 2: Apply the condition for a unique solution
For a system of linear equations to have a unique solution, the following condition must hold:
\[ \frac{a}{l} \neq \frac{b}{m} \]

Substituting the values we identified:
\[ \frac{\alpha}{12} \neq \frac{4}{\alpha} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ \alpha^2 \neq 12 \cdot 4 \]

### Step 4: Calculate the right-hand side
Calculating \( 12 \cdot 4 \):
\[ 12 \cdot 4 = 48 \]
Thus, we have:
\[ \alpha^2 \neq 48 \]

### Step 5: Solve for alpha
To find the values of \( \alpha \) that do not satisfy this condition, we take the square root of both sides:
\[ \alpha \neq \pm \sqrt{48} \]

### Step 6: Simplify the square root
We can simplify \( \sqrt{48} \):
\[ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \]
Thus, the values of \( \alpha \) that do not allow for a unique solution are:
\[ \alpha \neq 4\sqrt{3} \quad \text{and} \quad \alpha \neq -4\sqrt{3} \]

### Conclusion
The system of equations will have a unique solution for all values of \( \alpha \) except \( \alpha = 4\sqrt{3} \) and \( \alpha = -4\sqrt{3} \).