[tex]\boxed{x=1, \y=-1, \ z=2}[/tex]
We will use the Gaussian elimination method to solve this problem. To do so, let's follow the following steps:
Step 1: Let's multiply first equation by −2. Next, add the result to the second equation. So:
[tex]\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\~~ x&-~~~~~ y&+~~~~ z&~=~4\end{array}[/tex]
Step 2: Let's multiply first equation by −1. Next, add the result to the third equation. Thus:
[tex]\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&-~~~3~ y&+~~2~ z&~=~7\end{array}[/tex]
Step 3: Let's multiply second equation by −35, Next, add the result to the third equation. Therefore:
[tex]\begin{array}{ cccc }~~ x&+~~2~ y&-~~~~~ z&~=~-3\\&-~~~5~ y&+~~3~ z&~=~11\\&&+~~\frac{ 1 }{ 5 }~ z&~=~\frac{ 2 }{ 5 }\end{array}[/tex]
Step 4: solve for z, then for y, then for x:
[tex] \frac{ 1 }{ 5 } ~ z & = \frac{ 2 }{ 5 } \\ \\ \boxed{z & = 2}[/tex]
[tex]-5y+3z &= 11\\-5y+3\cdot 2 &= 11\\ \\ \boxed{y &= -1}[/tex]
By substituting [tex]y=-1 \ and \ z=2[/tex] into the first equation, we get the [tex]x[/tex]. So:
[tex]x+2(-1)-2=-3 \\ \\ x-2-2=-3 \\ \\ \boxed{x=1}[/tex]
