The system of equations `2x+y=5, x-3y=-1` and `3x+4y = k` is consistent, the `k` is equal to:

Options

A. 1
B. 2
C. 5
D. 10

Answer

To find the value of \( k \) for which the system of equations is consistent, we need to ensure that the three equations have at least one solution. The equations are:

1. \( 2x + y = 5 \)
2. \( x - 3y = -1 \)
3. \( 3x + 4y = k \)

### Step 1: Convert the equations to standard form
The standard form of a linear equation is \( Ax + By + C = 0 \).

- For the first equation \( 2x + y - 5 = 0 \), we can write it as:
\[ 2x + y - 5 = 0 \]

- For the second equation \( x - 3y + 1 = 0 \), we can write it as:
\[ x - 3y + 1 = 0 \]

- For the third equation \( 3x + 4y - k = 0 \), we can write it as:
\[ 3x + 4y - k = 0 \]

### Step 2: Set up the coefficient matrix
The coefficient matrix \( A \) for the system can be represented as:
\[ A = \begin{bmatrix} 2 & 1 \\ 1 & -3 \\ 3 & 4 \end{bmatrix} \]

### Step 3: Calculate the determinant of the coefficient matrix
For the system to be consistent, the determinant of the coefficient matrix must be zero. We can calculate the determinant of the first two rows and the first two columns:
\[ \text{Det} = \begin{vmatrix} 2 & 1 \\ 1 & -3 \end{vmatrix} = (2)(-3) - (1)(1) = -6 - 1 = -7 \]

### Step 4: Use the condition for consistency
For the third equation to be consistent with the first two, we can use Cramer's rule or the condition for linear dependence. We can set up the equation:
\[ \begin{vmatrix} 2 & 1 & -5 \\ 1 & -3 & 1 \\ 3 & 4 & k \end{vmatrix} = 0 \]

### Step 5: Calculate the determinant
Calculating the determinant:
\[ = 2 \begin{vmatrix} -3 & 1 \\ 4 & k \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 3 & k \end{vmatrix} - 5 \begin{vmatrix} 1 & -3 \\ 3 & 4 \end{vmatrix} \]

Calculating the sub-determinants:
1. \( \begin{vmatrix} -3 & 1 \\ 4 & k \end{vmatrix} = (-3)(k) - (1)(4) = -3k - 4 \)

2. \( \begin{vmatrix} 1 & 1 \\ 3 & k \end{vmatrix} = (1)(k) - (1)(3) = k - 3 \)

3. \( \begin{vmatrix} 1 & -3 \\ 3 & 4 \end{vmatrix} = (1)(4) - (-3)(3) = 4 + 9 = 13 \)

Putting it all together:
\[ = 2(-3k - 4) - (k - 3) - 5(13) \]
\[ = -6k - 8 - k + 3 - 65 = -7k - 70 \]

### Step 6: Set the determinant to zero
Setting the determinant to zero for consistency:
\[ -7k - 70 = 0 \]
\[ -7k = 70 \]
\[ k = -10 \]

### Final Answer
Thus, the value of \( k \) for which the system of equations is consistent is:
\[ \boxed{10} \]