If the line `(2+k)x-(2-k)y+(4k+14)=0` passes through the point (-1, 21), find k.

Answer

To solve the problem, we need to determine the value of \( k \) such that the line given by the equation

\[ (2+k)x - (2-k)y + (4k + 14) = 0 \]

passes through the point \((-1, 21)\).

### Step-by-Step Solution:

1. **Substitute the Point into the Line Equation**:
We know that if the point \((-1, 21)\) lies on the line, then substituting \( x = -1 \) and \( y = 21 \) into the equation should satisfy it. Thus, we substitute:

\[ (2+k)(-1) - (2-k)(21) + (4k + 14) = 0 \]

2. **Simplify the Equation**:
Now we simplify the left-hand side:

\[ -(2+k) - (2-k)(21) + (4k + 14) = 0 \]

Expanding \(-(2-k)(21)\):

\[ -(2+k) - (42 - 21k) + (4k + 14) = 0 \]

This simplifies to:

\[ -2 - k - 42 + 21k + 4k + 14 = 0 \]

3. **Combine Like Terms**:
Now we combine the constant terms and the terms involving \( k \):

- Constant terms: \(-2 - 42 + 14 = -30\)
- Terms with \( k \): \(-k + 21k + 4k = 24k\)

Thus, we have:

\[ 24k - 30 = 0 \]

4. **Solve for \( k \)**:
Rearranging gives us:

\[ 24k = 30 \]

Dividing both sides by 24:

\[ k = \frac{30}{24} = \frac{5}{4} \]

### Final Answer:
The value of \( k \) is

\[ \boxed{\frac{5}{4}} \]