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System of equations x-y-2z=4, -x+2y+z=1, x+y-3z=11. Solve by elimination

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[tex]x-y-2z=4 \\ -x+2y+z=1 \\ x+y-3z=11 \\ \\ \left \{ {{x-y-2z=4 } \atop {-x+2y+z=1}} \right \ and \ \left \{ {{-x+2y+z=1 } \atop {x+y-3z=11}} \right \\ \\ \hbox{1st system:} \\ x-y-2z=4 \\ \underline{-x+2y+z=1} \\ x-x+2y-y-2z+z=4+1 \\ y-z=5[/tex]

[tex]\hbox{2nd system:} \\ -x+2y+z=1 \\ \underline{x+y-3z=11} \\ x-x+2y+y-3z+z=1+11 \\ 3y-2z=12 \\ \\ \hbox{now compare both:} \\ y-z=5 \ \ \ |\cdot (-2) \\ 3y-2z=12 \\ \\ -2y+2z=-10 \\ \underline{3y-2z=12 \ \ \ \ \ \ } \\ 3y-2y+2z-2z=12-10 \\ y=2 \\ \\ 2-z=5 \\ -z=3 \\ z=-3 \\ \\ x-2-2 \cdot (-3) = 4 \\ x-2+6=4 \\ x+4=4 \\ x=0[/tex]

The answer is: x=0, y=2, z=-3.
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