The value of k for which the system of equations x+y -4 = 0 and 2x + ky = 3 has no solution, is:

Options

A. `-2`
B. `ne 2`
C. 2
D. 3

Answer

To find the value of \( k \) for which the system of equations has no solution, we can follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( x + y - 4 = 0 \)
2. \( 2x + ky = 3 \)

We can rewrite them in the standard form \( Ax + By + C = 0 \):
1. \( x + y - 4 = 0 \) can be rewritten as \( 1x + 1y - 4 = 0 \).
2. \( 2x + ky - 3 = 0 \) can be rewritten as \( 2x + ky - 3 = 0 \).

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation: \( A_1 = 1, B_1 = 1, C_1 = -4 \)
- For the second equation: \( A_2 = 2, B_2 = k, C_2 = -3 \)

### Step 3: Apply the condition for no solution
For the system of equations to have no solution, the following condition must hold:
\[ \frac{A_1}{B_1} = \frac{A_2}{B_2} \quad \text{and} \quad \frac{A_1}{B_1} \neq \frac{C_1}{C_2} \]

Substituting the values:
\[ \frac{1}{1} = \frac{2}{k} \]
This simplifies to:
\[ 1 = \frac{2}{k} \]

### Step 4: Solve for \( k \)
Cross-multiplying gives:
\[ k = 2 \]

### Step 5: Check the second condition
Now we check the second condition:
\[ \frac{C_1}{C_2} = \frac{-4}{-3} = \frac{4}{3} \]
We need to ensure:
\[ 1 \neq \frac{4}{3} \]
This condition is satisfied since \( 1 \) is not equal to \( \frac{4}{3} \).

### Conclusion
Thus, the value of \( k \) for which the system of equations has no solution is:
\[ \boxed{2} \]

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