For what value of k, the system of equations 3x - 2y = 5 and 6x - ky = 8 has no unique solution?

Answer

To determine the value of \( k \) for which the system of equations has no unique solution, we need to analyze the conditions under which two linear equations are either parallel or coincident. A system of linear equations has no unique solution when the lines represented by the equations are parallel.

The given equations are:
1. \( 3x - 2y = 5 \) (Equation 1)
2. \( 6x - ky = 8 \) (Equation 2)

### Step 1: Rewrite the equations in the standard form
We can rewrite both equations in the form \( Ax + By + C = 0 \).

From Equation 1:
\[ 3x - 2y - 5 = 0 \]

From Equation 2:
\[ 6x - ky - 8 = 0 \]

### Step 2: Identify the coefficients
From the rewritten equations, we can identify the coefficients:
- For Equation 1: \( A_1 = 3, B_1 = -2, C_1 = -5 \)
- For Equation 2: \( A_2 = 6, B_2 = -k, C_2 = -8 \)

### Step 3: Set up the condition for no unique solution
For the system to have no unique solution, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \]

Substituting the coefficients:
\[ \frac{3}{6} = \frac{-2}{-k} \]

### Step 4: Simplify the ratios
Simplifying the left side:
\[ \frac{1}{2} = \frac{2}{k} \]

### Step 5: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 1 \cdot k = 2 \cdot 2 \]
\[ k = 4 \]

### Conclusion
The value of \( k \) for which the system of equations has no unique solution is \( k = 4 \).

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