Find the value of k such that the line
`(k-2)x+(k+3)y-5=0` is
parallel to the line `2x-y+7=0`

Answer

To find the value of \( k \) such that the line \( (k-2)x + (k+3)y - 5 = 0 \) is parallel to the line \( 2x - y + 7 = 0 \), we will follow these steps:

### Step 1: Identify the slopes of both lines
The slope of a line in the form \( Ax + By + C = 0 \) can be found using the formula:
\[ m = -\frac{A}{B} \]

For the first line \( (k-2)x + (k+3)y - 5 = 0 \):
- Here, \( A = k - 2 \) and \( B = k + 3 \).
- Thus, the slope \( m_1 \) is:
\[ m_1 = -\frac{k-2}{k+3} \]

For the second line \( 2x - y + 7 = 0 \):
- Here, \( A = 2 \) and \( B = -1 \).
- Thus, the slope \( m_2 \) is:
\[ m_2 = -\frac{2}{-1} = 2 \]

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

### Step 3: Solve for \( k \)
To solve the equation, we first eliminate the negative sign:
\[ \frac{k-2}{k+3} = -2 \]
Cross-multiplying gives:
\[ k - 2 = -2(k + 3) \]
Expanding the right side:
\[ k - 2 = -2k - 6 \]
Now, we will move all terms involving \( k \) to one side and constant terms to the other side:
\[ k + 2k = -6 + 2 \]
This simplifies to:
\[ 3k = -4 \]
Now, dividing both sides by 3:
\[ k = -\frac{4}{3} \]

### Conclusion
Thus, the value of \( k \) such that the line \( (k-2)x + (k+3)y - 5 = 0 \) is parallel to the line \( 2x - y + 7 = 0 \) is:
\[ \boxed{-\frac{4}{3}} \]