For what value of k does the system of equation `3x+ky=11` and `5x-7y=6` has no solution?

Answer

To determine the value of \( k \) for which the system of equations

\[ 3x + ky = 11 \quad (1) \]
\[ 5x - 7y = 6 \quad (2) \]

has no solution, we need to analyze the conditions under which two linear equations do not intersect. This occurs when the lines represented by the equations are parallel. For two lines to be parallel, the ratios of their coefficients must be equal, but the ratio of the constants must not be equal.

### Step 1: Identify coefficients and constants
From the equations, we can identify the coefficients and constants:
- For equation (1):
- \( a_1 = 3 \)
- \( b_1 = k \)
- \( c_1 = -11 \) (after moving 11 to the left side)

- For equation (2):
- \( a_2 = 5 \)
- \( b_2 = -7 \)
- \( c_2 = -6 \) (after moving 6 to the left side)

### Step 2: Set up the ratio condition for no solution
The condition for no solution is given by:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]

Substituting the values we have:

\[ \frac{3}{5} = \frac{k}{-7} \quad (3) \]

### Step 3: Solve for \( k \)
To find \( k \), we can cross-multiply:

\[ 3 \cdot (-7) = 5 \cdot k \]
\[ -21 = 5k \]

Now, divide both sides by 5:

\[ k = \frac{-21}{5} \]

### Step 4: Verify the condition for no solution
We need to check that the second condition holds, which is:

\[ \frac{3}{5} \neq \frac{-11}{-6} \]

Calculating \( \frac{-11}{-6} \):

\[ \frac{-11}{-6} = \frac{11}{6} \]

Now, we compare:

\[ \frac{3}{5} \neq \frac{11}{6} \]

This is true, confirming that the lines are indeed parallel when \( k = \frac{-21}{5} \).

### Final Answer
The value of \( k \) for which the system of equations has no solution is:

\[ \boxed{-\frac{21}{5}} \]