Precalculus (6th Edition) Blitzer

The solution of the system is $\left( -1,-2,3 \right)$.
Now, it is given that $2x+3y+7z=13$ ...... (I) $3x+2y-5z=-2z$ ...... (II) $5x+7y-3z=-28$ ...... (III) Multiply equation (I) by 3 and equation (II) by 2 and subtract: \begin{align} & 6x+9y+21z=39 \\ & 6x+4y-10z=-44 \\ \end{align} Solve and we get: $5y+31z=83$ ...... (IV) Now, multiply equation (II) by 5 and equation (III) by 3 and subtract: \begin{align} & 15x+10y-25z=-110 \\ & 15x+21y-9z=-84 \\ & -11y-16z=-26 \end{align} Solve and we get: $11y+16z=26$ ...... (V) Multiply equation (IV) by 11 and equation (V) by 5 and subtract: \begin{align} & 55y+341z=913 \\ & 55y+60z=130 \\ & 261z=783 \end{align} Solve and we get: $z=3$ Putting the value of $z=3$ in equation (V): \begin{align} & 11y+16\left( 3 \right)=26 \\ & 11y+48=26 \\ & 11y=-22 \\ & y=-2 \end{align} Putting the value of $z=3$ and $y=-2$ in equation (I): \begin{align} & 2x-6+21=13 \\ & 2x=13-15 \\ & x=-1 \end{align} Thus, the ordered triple $\left( -1,-2,3 \right)$ satisfies the three systems.