Answer:
Required unique solutions are 8, 4, 0 and the system is consistant.
Explanation:
Given system of equation,
5x-2y+4z=12
x+y+z=8
4x-3y+3z=4
To find solution of the system re-write as,
![\left[\begin{array}{ccc}5&-2&4 :12\\1&1&1 : 8\\4&-3&3 : 4\end{array}\right]\to\left[\begin{array}{ccc}1&1&1 :8\\5&-2&4 : 12\\4&-3&3 : 4\end{array}\right] (R_1\iff R_2)\\\implies \left[\begin{array}{ccc}1&1& 1:8\\0&-7&-1 : -28\\0&-7&-1 : -28\end{array}\right] (R_(2)^(')=R_2-5R_1,R_(3)^(')=R_3-4R_1)\\\implies \left[\begin{array}{ccc}1&1&1 :8\\0&-7&-1 :-28\\0&0&0 : 0\end{array}\right] (R_(3)^(')=R_3+R_2)\\\implies \left[\begin{array}{ccc}1&1&1 :8\\0&1&1/7 : 4\\0&0&0 : 0\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/39g4mdqin8jbh14686ifu4jf6za9newqsn.png)
Hence solutions are 8, 4 and 0. And it is a unique solution. Therefore system is consistent.