[tex]x=-2 \\ \\ y=6 \\ \\ z=4[/tex]
We have the following System of Linear Equations in three variables:
[tex]\begin{array}{c}(1)\\(2)\\(3)\end{array}\left\{ \begin{array}{c}3x+3y+z=16\\x-3y+2z=-12\\8x-2y+3z=-16\end{array}\right.[/tex]
So let's solve it step by step using elimination method:
Step 1: Swap Row 1 and Row 2. So we have:
[tex]\begin{array}{ cccc }~~ x&-~~~3~ y&+~~2~ z&~=~-12\\3~ x&+~~3~ y&+~~~~ z&~=~16\\8~ x&-~~~2~ y&+~~3~ z&~=~-16\end{array}[/tex]
Step 2: Multiply first equation by [tex]-3[/tex] and add the result to the second equation. So we get:
[tex]\begin{array}{ cccc }~~ x&-~~~3~ y&+~~2~ z&~=~-12\\&+~~12~ y&-~~~5~ z&~=~52\\8~ x&-~~~2~ y&+~~3~ z&~=~-16\end{array}[/tex]
Step 3: Multiply first equation by [tex]-8[/tex] and add the result to the third equation. So we get:
[tex]\begin{array}{ cccc }~~ x&-~~~3~ y&+~~2~ z&~=~-12\\&+~~12~ y&-~~~5~ z&~=~52\\&+~~22~ y&-~~~13~ z&~=~80\end{array}[/tex]
Step 4: Multiply second equation by [tex]-11/6[/tex] and add the result to the third equation. So we get:
[tex]\begin{array}{ cccc }~~ x&-~~~3~ y&+~~2~ z&~=~-12\\&+~~12~ y&-~~~5~ z&~=~52\\&&-~~~\frac{ 23 }{ 6 }~ z&~=~-\frac{ 46 }{ 3 }\end{array}[/tex]
Step 5: solve for [tex]z[/tex].
[tex]\begin{aligned} -\frac{ 23 }{ 6 } ~ z & = -\frac{ 46 }{ 3 } \\ z & = 4 \end{aligned}[/tex]
Step 6: solve for [tex]y[/tex]:
[tex]\begin{aligned}12y-5z &= 52\\12y-5\cdot 4 &= 52\\y &= 6 \end{aligned}[/tex]
Step 7: solve for [tex]x[/tex] by substituting [tex]y=6[/tex] and [tex]z=4[/tex] into the first equation:
[tex]3x+3(6)+(4)=16 \\ \\ 3x+18+4=16 \\ \\ 3x+22=16 \\ \\ 3x=16-22 \\ \\ 3x=-6 \\ \\ x=-\frac{6}{3} \\ \\ x=-2[/tex]
The substitution method: https://brainly.com/question/10852714
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