Answer
$\{(\frac{2z+14}{23},\frac{7z-6}{23},z)\}$
Work Step by Step
Use Cramer's rule, we have:
1. $D=\begin{vmatrix} 7 & 1 & -1 \\ 2 & -3 & 1 \\ -6 & 9 & -3 \end{vmatrix}=7((-3)(-3)-(1)(9))-2((1)(-3)-(-1)(9))-6((1)(1)-(-1)(-3))=0$
2. Since $D=0$, we can not use the Cramer's rule and the system has unlimited solutions.
3. We can see that the second and the third equations are the same.
4. Rewrite the first two equations as $\begin{cases} 7x+y=4+z \\ 2x-3y=2-z \end{cases}$ then multiply 3 to the first equation and add the result to the second to get $23x=14+2z$ which gives $x=\frac{2z+14}{23}$
5. We can find $y=4+z-7x=4+z-7(\frac{2z+14}{23})=\frac{7z-6}{23}$
6. Thus the solution set is $\{(\frac{2z+14}{23},\frac{7z-6}{23},z)\}$ (one can also write the solutions in terms of $x$ or $y$)
