## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( \frac{1}{2},\frac{1}{3},-1 \right)$.
It is known that: $3\left( 2x+y \right)+5z=-1$ ...... (I) $2\left( x-3y+4z \right)=-9$ ...... (II) $4\left( 1+x \right)=-3\left( z-3y \right)$ ...... (III) From equation (I): $3\left( 2x+y \right)+5z=-1$ $6x+3y+5z=-1$ ...... (IV) From equation (II): $2\left( x-3y+4z \right)=-9$ $2x-6y+8z=-9$ ...... (V) From equation (III): \begin{align} & 4\left( 1+x \right)=-3\left( z-3y \right) \\ & 4+4x=-3z+9y \end{align} $4x-9y+3z=-4$ ...... (VI) Multiply equation (V) by 3 and subtract from equation (IV): \begin{align} & 6x+3y+5z=-1 \\ & 6x-18y+24z=-27 \\ \end{align} $21y-192=26$ ...... (VII) Now multiply equation (V) by 7 and subtract from equation (VII): \begin{align} & 21y-19z=26 \\ & 21y-91z=98 \\ \end{align} $3y-13z=14$ ...... (VIII) Multiply equation (VIII) by 7 and subtract from equation (VII): \begin{align} & 21y-19z=26 \\ & 21y-91z=98 \\ \end{align} Solve and get: \begin{align} & 72z=-72 \\ & z=-1 \end{align} Put the value of z in equation (VII) \begin{align} & 21y+19=26 \\ & 21y=7 \\ & y=\frac{1}{3} \end{align} Put the value of $y=\frac{1}{3}$ and $z=-1$ in equation (IV): \begin{align} & 6x+3\left( \frac{1}{3} \right)-5=-1 \\ & x=\frac{1}{2} \end{align} Hence, the ordered triple $\left( \frac{1}{2},\frac{1}{3},-1 \right)$ satisfies the three systems.