Solve the following system of equations by elimination:

1. \( x - y - 2z = 4 \)
2. \( -x + 2y + z = 1 \)
3. \( x + y - 3z = 11 \)

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
2nd system: − x + 2 y + z = 1 x + y − 3 z = 11 ​ x − x + 2 y + y − 3 z + z = 1 + 11 3 y − 2 z = 12 now compare both: y − z = 5 ∣ ⋅ ( − 2 ) 3 y − 2 z = 12 − 2 y + 2 z = − 10 3 y − 2 z = 12 ​ 3 y − 2 y + 2 z − 2 z = 12 − 10 y = 2 2 − z = 5 − z = 3 z = − 3 x − 2 − 2 ⋅ ( − 3 ) = 4 x − 2 + 6 = 4 x + 4 = 4 x = 0
The answer is: x=0, y=2, z=-3.

Answered by naǫ | 2024-06-10

The solutions to the system of equations are x=0, y=2, and z=-3. This was achieved by using the elimination method to remove variables systematically. The final values were derived through substituting back into the original equations.
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Answered by naǫ | 2024-10-18