systems with three variables

[KeannaCoolBeans]

So ive done this like 3 times and i still never get the right answer (( HELPPP!!!!
3x + 2y - z = 12
-4x + y - 2z = 4
x - 3y + z = -4

 

[soroban]

Hello, KeannaCoolBeans!

\(\displaystyle \begin{array}{cccc} 3x + 2y - z &=& 12 & [1] \\ \text{-}4x + y - 2z &=& 4 & [2] \\ x - 3y + z &=& \text{-}4 & [3] \end{array}\)

\(\displaystyle \text{Add [1] and [3]: }\;\begin{array}{ccc}3x + 2y - z &=& 12 \\ x - 3y + z &=& \text{-}4 \\ \hline 4x - y \qquad &=& 8 \end{array}\)

. . \(\displaystyle \begin{array}{ccccccc}\text{Add [2]} \\ \text{and }\text{-}2\!\cdot\![1]} \\ \\ \end{array}\;\; \begin{array}{cccccc}\text{-}4x + y - 2z &=& 4 \\ \text{-}6x -4y + 2x &=&\text{-}24 \\ \hline \text{-}10x - 3y \qquad &=& \text{-}20 \end{array}\)


We have: .\(\displaystyle \begin{array}{cccc}4x - y &=& 8 & [4] \\ 10x + 3y &=& 20 & [5] \end{array}\)

\(\displaystyle \begin{array}{c}\text{Add [5]} \\ \text{and }3\!\cdot\![4]} \\ \\ \end{array}\;\; \begin{array}{ccccc}10x + 3y &=& 20 \\ 12x - 3y &=& 24 \\ \hline 22x \qquad &=& 44\end{array}\)

Hence, we have: .\(\displaystyle \boxed{x \,=\, 2}\)

Substitute into [4]: .\(\displaystyle 4(2) - y \:=\:8 \quad\Rightarrow\quad \boxed{y \,=\,0}\)

Substitute into [3]: .\(\displaystyle 2 - 3(0) + z \:=\:\text{-}4 \quad\Rightarrow\quad \boxed{z \,=\,\text{-}6}\)