Solve the 3 × 3 system shown below. Enter the values of x, y, and z. X + 2y – z = –3 (1) 2x – y + z = 5 (2) x – y + z = 4 (3)

Answer:[tex]\boxed{x=1, \y=-1, \ z=2}[/tex]Step-by-step explanation: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]