`k=a-b+12,k=b-c-17,k=c-a+11`
In the system of equations above, what is the value of k?

Answer

To solve the system of equations given by:

1. \( k = a - b + 12 \)
2. \( k = b - c - 17 \)
3. \( k = c - a + 11 \)

we will follow these steps:

### Step 1: Write down the equations
We have three equations:
- \( k = a - b + 12 \) (Equation 1)
- \( k = b - c - 17 \) (Equation 2)
- \( k = c - a + 11 \) (Equation 3)

### Step 2: Set up the equation for summation
We can add all three equations together:
\[ k + k + k = (a - b + 12) + (b - c - 17) + (c - a + 11) \]

### Step 3: Simplify the left side
The left side simplifies to:
\[ 3k \]

### Step 4: Simplify the right side
Now, let's simplify the right side:
- Combine like terms:
\[ (a - b + b - c + c - a) + (12 - 17 + 11) \]
- The terms \( a \), \( -b \), \( b \), \( -c \), \( c \), and \( -a \) cancel out:
\[ 0 + (12 - 17 + 11) \]
- Now calculate the constants:
\[ 12 - 17 = -5 \]
\[ -5 + 11 = 6 \]
So, the right side simplifies to:
\[ 6 \]

### Step 5: Set the equations equal
Now we have:
\[ 3k = 6 \]

### Step 6: Solve for k
To find \( k \), divide both sides by 3:
\[ k = \frac{6}{3} = 2 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{2} \]

---