If equation of two lines are `3x+8y=11` and `6x-ky=22` has infinite solution. Find value of k.

Answer

To find the value of \( k \) such that the equations \( 3x + 8y = 11 \) and \( 6x - ky = 22 \) have infinite solutions, we follow these steps:

### Step 1: Understand the condition for infinite solutions
Two lines will have infinite solutions if they are coincident, which means they represent the same line. This occurs when the ratios of the coefficients of \( x \), \( y \), and the constant terms are equal.

### Step 2: Write the equations in standard form
The given equations are:
1. \( 3x + 8y = 11 \)
2. \( 6x - ky = 22 \)

We can rewrite these equations in the form \( Ax + By + C = 0 \):
1. \( 3x + 8y - 11 = 0 \) (Here, \( A_1 = 3, B_1 = 8, C_1 = -11 \))
2. \( 6x - ky + 22 = 0 \) (Here, \( A_2 = 6, B_2 = -k, C_2 = 22 \))

### Step 3: Set up the ratios
For the lines to be coincident, we need:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]
Substituting the values we have:
\[ \frac{3}{6} = \frac{8}{-k} = \frac{-11}{22} \]

### Step 4: Simplify the ratios
1. Simplifying \( \frac{3}{6} \):
\[ \frac{3}{6} = \frac{1}{2} \]

2. Simplifying \( \frac{-11}{22} \):
\[ \frac{-11}{22} = -\frac{1}{2} \]

### Step 5: Set up the equation for \( k \)
Now we can equate \( \frac{3}{6} \) and \( \frac{8}{-k} \):
\[ \frac{1}{2} = \frac{8}{-k} \]

### Step 6: Cross-multiply to solve for \( k \)
Cross-multiplying gives:
\[ 1 \cdot (-k) = 2 \cdot 8 \]
\[ -k = 16 \]

### Step 7: Solve for \( k \)
Multiplying both sides by -1:
\[ k = -16 \]

### Final Answer
Thus, the value of \( k \) is \( -16 \).

---