Obtain an equivalent system by performing the stated elementary operation on the system. Replace the fourth equation by the sum of itself and -2 times the first equation. \[ \begin{array}{rr} 6 x+15 y-5 z= & 30 \\ 3 x+15 y-14 z= & -15 \\ 4 x-8 y-2 z= & 16 \\ 5 x+14 y+8 z= & -20 \end{array} \] \[ \begin{array}{l} \begin{array}{l} 6 x+15 y-5 z= \\ 3 x+15 y-14 z= \\ 3 x-15 \\ 4 x-8 y-2 z= \\ 5 x+16 \end{array} \rightarrow \begin{array}{l} \square x+\square y-\square z=\square \\ 5 x+14 y+8 z= \end{array} \rightarrow-20 \end{array} \begin{array}{l} \square x-\square y-\square z=\square \\ \square x-\square y+\square z=\square \end{array} \] (Type integers or simplified fractions.)

Let’s modify the fourth equation by replacing it with the sum of itself and -2 times the first equation. The first equation is: \[ 6x + 15y - 5z = 30 \] Multiplying this by -2, we get: \[ -12x - 30y + 10z = -60 \] Now, adding it to the fourth equation: \[ 5x + 14y + 8z = -20 \] Calculating: \[ (5x - 12x) + (14y - 30y) + (8z + 10z) = -20 - 60 \] This simplifies to: \[ -7x - 16y + 18z = -80 \] Thus, the new equivalent system of equations is: \[ \begin{array}{rr} 6 x+15 y-5 z= & 30 \\ 3 x+15 y-14 z= & -15 \\ 4 x-8 y-2 z= & 16 \\ -7 x-16 y+18 z= & -80 \end{array} \] The third equation remains unchanged as it was not part of the alteration process. Here is the final equivalent system: \[ \begin{array}{rr} 6 x + 15 y - 5 z = & 30 \\ 3 x + 15 y - 14 z = & -15 \\ 4 x - 8 y - 2 z = & 16 \\ -7 x - 16 y + 18 z = & -80 \end{array} \]