Find the solution of the system of equations. \[ \begin{array}{c}8 x+14 y=2 \\ -3 x-7 y=8\end{array} \]

To solve the system of equations: \[ \begin{array}{c} 8x + 14y = 2 \quad (1) \\ -3x - 7y = 8 \quad (2) \end{array} \] We can use the elimination or substitution method. Here, let's use the elimination method. First, we can manipulate the equations to align the coefficients for elimination. Multiply equation (2) by 2 so that the coefficients of \(y\) match: \[ -6x - 14y = 16 \quad (3) \] Now, we have the transformed system: \[ \begin{array}{c} 8x + 14y = 2 \quad (1) \\ -6x - 14y = 16 \quad (3) \end{array} \] Now, we can add equations (1) and (3): \[ (8x + 14y) + (-6x - 14y) = 2 + 16 \] This simplifies to: \[ 2x = 18 \] Dividing both sides by 2 gives: \[ x = 9 \] Now substitute \(x = 9\) back into equation (1) to find \(y\): \[ 8(9) + 14y = 2 \] This simplifies to: \[ 72 + 14y = 2 \] Subtract 72 from both sides: \[ 14y = 2 - 72 \] \[ 14y = -70 \] Now divide by 14: \[ y = -5 \] Thus, the solution to the system of equations is: \[ (x, y) = (9, -5) \]