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Solve the system of equations by substitution.\newline−2x+y−z=−8-2x + y - z = -8\newline−3x−3y−3z=−9-3x - 3y - 3z = -9\newlinez=−2z = -2

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Q. Solve the system of equations by substitution.\newline−2x+y−z=−8-2x + y - z = -8\newline−3x−3y−3z=−9-3x - 3y - 3z = -9\newlinez=−2z = -2
  1. Solve for y: Solve for y using the first equation by substituting z=−2z = -2.\newline−2x+y−(−2)=−8-2x + y - (-2) = -8\newline−2x+y+2=−8-2x + y + 2 = -8\newliney=−8−2+2xy = -8 - 2 + 2x\newliney=−10+2xy = -10 + 2x
  2. Substitute y and z: Substitute y and z into the second equation.\newline−3x−3(−10+2x)−3(−2)=−9-3x - 3(-10 + 2x) - 3(-2) = -9\newline−3x+30−6x+6=−9-3x + 30 - 6x + 6 = -9\newlineCombine like terms.\newline−9x+36=−9-9x + 36 = -9
  3. Combine like terms: Solve for xx.−9x=−9−36-9x = -9 - 36−9x=−45-9x = -45x=−45−9x = \frac{-45}{-9}x=5x = 5
  4. Solve for xx: Substitute xx back into the equation for yy.y=−10+2(5)y = -10 + 2(5)y=−10+10y = -10 + 10y=0y = 0
  5. Substitute xx back: We already have zz from the third equation.z=−2z = -2

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