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Solve the system of equations by elimination.\newline2x−2y−z=−62x - 2y - z = -6\newline−x−3y+3z=−6-x - 3y + 3z = -6\newline−x+y+z=6-x + y + z = 6

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Solve a system of equations in three variables using eliminationright chevron icon

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Q. Solve the system of equations by elimination.\newline2x−2y−z=−62x - 2y - z = -6\newline−x−3y+3z=−6-x - 3y + 3z = -6\newline−x+y+z=6-x + y + z = 6
  1. Multiply by 22: First, let's multiply the third equation by 22 so we can eliminate zz with the first equation.\newline−2(−x+y+z)=2(6)-2(-x + y + z) = 2(6)\newline2x−2y−2z=122x - 2y - 2z = 12
  2. Add to eliminate z: Now, add the modified third equation to the first equation to eliminate z.\newline(2x−2y−z)+(2x−2y−2z)=−6+12(2x - 2y - z) + (2x - 2y - 2z) = -6 + 12\newline4x−4y−3z=64x - 4y - 3z = 6\newlineBut wait, there's a mistake here, I didn't eliminate z correctly.

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