Given the system of equations, what is the value of the \( y \)-determinant? \( \begin{array}{l}3 x+y-10=0 \\ 4 x-y-4=0 \\ -14 \\ 28 \\ -28\end{array} \)

To find the value of the \( y \)-determinant for the given system of equations, we first need to express the system in a standard form and then calculate the determinant of the coefficient matrix. The given equations are: 1. \( 3x + y - 10 = 0 \) 2. \( 4x - y - 4 = 0 \) We can rewrite these equations in the standard form \( Ax + By = C \): 1. \( 3x + y = 10 \) 2. \( 4x - y = 4 \) Now, we can represent this system in matrix form \( A \mathbf{x} = \mathbf{b} \), where: \[ A = \begin{pmatrix} 3 & 1 \\ 4 & -1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 10 \\ 4 \end{pmatrix} \] Next, we will calculate the determinant of the coefficient matrix \( A \): \[ \text{det}(A) = \begin{vmatrix} 3 & 1 \\ 4 & -1 \end{vmatrix} \] The determinant can be calculated using the formula: \[ \text{det}(A) = ad - bc \] where \( a = 3, b = 1, c = 4, d = -1 \). Now, let's calculate the determinant: \[ \text{det}(A) = (3)(-1) - (1)(4) = -3 - 4 = -7 \] Thus, the value of the \( y \)-determinant is \(-7\).