The equations (a-3)x + 3y = 12 and 4x+(a -2y) = k have infinitely many solutions. What is the sum of all the possible vlaue of k ?

Answer

To solve the problem, we need to find the values of \( k \) such that the two equations have infinitely many solutions. The equations given are:

1. \((a - 3)x + 3y = 12\)
2. \(4x + (a - 2)y = k\)

### Step 1: Set up the condition for infinitely many solutions
For the two equations to have infinitely many solutions, the ratios of the coefficients of \( x \), \( y \), and the constants must be equal. This gives us the following condition:

\[ \frac{a - 3}{4} = \frac{3}{a - 2} = \frac{12}{k} \]

### Step 2: Set up the first ratio
From the first part of the ratio, we can write:

\[ \frac{a - 3}{4} = \frac{3}{a - 2} \]

Cross-multiplying gives:

\[ (a - 3)(a - 2) = 4 \cdot 3 \]

This simplifies to:

\[ (a - 3)(a - 2) = 12 \]

### Step 3: Expand and rearrange the equation
Expanding the left side:

\[ a^2 - 2a - 3a + 6 = 12 \]

This simplifies to:

\[ a^2 - 5a + 6 - 12 = 0 \]

Thus, we have:

\[ a^2 - 5a - 6 = 0 \]

### Step 4: Factor the quadratic equation
Now we will factor the quadratic:

\[ (a - 6)(a + 1) = 0 \]

This gives us the solutions:

\[ a = 6 \quad \text{or} \quad a = -1 \]

### Step 5: Find the corresponding values of \( k \)
Now we will substitute these values of \( a \) back into the ratio to find \( k \).

Using the ratio \(\frac{12}{k} = \frac{3}{a - 2}\):

1. For \( a = 6 \):

\[ \frac{12}{k} = \frac{3}{6 - 2} = \frac{3}{4} \]

Cross-multiplying gives:

\[ 12 \cdot 4 = 3k \implies 48 = 3k \implies k = \frac{48}{3} = 16 \]

2. For \( a = -1 \):

\[ \frac{12}{k} = \frac{3}{-1 - 2} = \frac{3}{-3} = -1 \]

Cross-multiplying gives:

\[ 12 = -k \implies k = -12 \]

### Step 6: Sum the possible values of \( k \)
Now we have two possible values of \( k \):

- \( k = 16 \)
- \( k = -12 \)

The sum of all possible values of \( k \) is:

\[ 16 + (-12) = 4 \]

### Final Answer
The sum of all the possible values of \( k \) is \( \boxed{4} \).