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

Answer

To find the value of \( k \) such that the system of equations \( 3x - 2y = 5 \) and \( 6x - ky = 8 \) has no unique solution, we need to analyze the conditions under which a system of linear equations does not have a unique solution.

### Step 1: Write the equations in standard form
We start with the given equations:
1. \( 3x - 2y = 5 \)
2. \( 6x - ky = 8 \)

We can rewrite these equations in the standard form \( Ax + By + C = 0 \):
1. \( 3x - 2y - 5 = 0 \) (Equation 1)
2. \( 6x - ky - 8 = 0 \) (Equation 2)

### Step 2: Identify coefficients
From the standard form, we 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 following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \]

Substituting the values we identified:
\[ \frac{3}{6} = \frac{-2}{-k} \]

### Step 4: Simplify the equation
Now we simplify 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
Thus, the value of \( k \) for which the system of equations has no unique solution is:
\[ \boxed{4} \]