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 solve for the \( y \)-determinant in the given system of equations, we first convert the equations into standard form: 1. From \( 3x + y - 10 = 0 \), we can express \( y \) as \( y = 10 - 3x \). 2. From \( 4x - y - 4 = 0 \), we can express \( y \) in this form as \( y = 4x - 4 \). Next, we set these two expressions for \( y \) equal to each other: \[ 10 - 3x = 4x - 4 \] Rearranging this gives: \[ 10 + 4 = 4x + 3x \] This simplifies to: \[ 14 = 7x \] Thus, we find: \[ x = 2 \] Now, we'll substitute \( x = 2 \) back into one of the equations to find \( y \): Using \( y = 4x - 4 \): \[ y = 4(2) - 4 = 8 - 4 = 4 \] The solution is \( x = 2 \) and \( y = 4 \). To find the \( y \)-determinant associated with the equations’ coefficients, we can represent the coefficients in a matrix and apply the determinant formula: The matrix is: \[ \begin{pmatrix} 3 & 1 \\ 4 & -1 \end{pmatrix} \] Calculating the determinant: \[ \text{Determinant} = (3)(-1) - (1)(4) = -3 - 4 = -7 \] So the absolute value of the \( y \)-determinant is \( 7 \).