It `2x+3y=19` and `kx+6y=39` are parallel lines find k.

Answer

To find the value of \( k \) such that the lines \( 2x + 3y = 19 \) and \( kx + 6y = 39 \) are parallel, we follow these steps:

### Step 1: Find the slope of the first line
The first line is given by the equation:
\[ 2x + 3y = 19 \]
To find the slope, we need to rearrange this equation into the slope-intercept form \( y = mx + b \).

1. Subtract \( 2x \) from both sides:
\[ 3y = -2x + 19 \]
2. Divide every term by \( 3 \):
\[ y = -\frac{2}{3}x + \frac{19}{3} \]

The slope \( m_1 \) of the first line is:
\[ m_1 = -\frac{2}{3} \]

### Step 2: Find the slope of the second line
The second line is given by the equation:
\[ kx + 6y = 39 \]
We will also rearrange this into slope-intercept form.

1. Subtract \( kx \) from both sides:
\[ 6y = -kx + 39 \]
2. Divide every term by \( 6 \):
\[ y = -\frac{k}{6}x + \frac{39}{6} \]

The slope \( m_2 \) of the second line is:
\[ m_2 = -\frac{k}{6} \]

### Step 3: Set the slopes equal to each other
Since the lines are parallel, their slopes must be equal:
\[ m_1 = m_2 \]
Substituting the values we found:
\[ -\frac{2}{3} = -\frac{k}{6} \]

### Step 4: Solve for \( k \)
To eliminate the negative signs, we can multiply both sides by -1:
\[ \frac{2}{3} = \frac{k}{6} \]

Next, we cross-multiply to solve for \( k \):
\[ 2 \cdot 6 = 3 \cdot k \]
\[ 12 = 3k \]
Now, divide both sides by \( 3 \):
\[ k = \frac{12}{3} = 4 \]

### Conclusion
The value of \( k \) is:
\[ \boxed{4} \]