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Solve the system of equations by elimination.\newline2x−y+z=−142x - y + z = -14\newlinex−y+z=−14x - y + z = -14\newline3x+y+3z=−103x + y + 3z = -10

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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−y+z=−142x - y + z = -14\newlinex−y+z=−14x - y + z = -14\newline3x+y+3z=−103x + y + 3z = -10
  1. Eliminate y and z: Subtract the second equation from the first to eliminate y and z.\newline(2x−y+z)−(x−y+z)=−14−(−14)(2x - y + z) - (x - y + z) = -14 - (-14)\newline2x−x=02x - x = 0\newlinex=0x = 0
  2. Find yy: Substitute x=0x = 0 into the second equation to find yy.0−y+z=−140 - y + z = -14−y+z=−14-y + z = -14
  3. Find zz: Substitute x=0x = 0 into the third equation to find zz.3(0)+y+3z=−103(0) + y + 3z = -10y+3z=−10y + 3z = -10
  4. Eliminate y: Now we have two equations with y and z:\newline−y+z=−14-y + z = -14\newliney+3z=−10y + 3z = -10\newlineAdd these two equations to eliminate y.\newline(−y+z)+(y+3z)=−14+(−10)(-y + z) + (y + 3z) = -14 + (-10)\newline4z=−244z = -24\newlinez=−6z = -6
  5. Find yy: Substitute z=−6z = -6 into −y+z=−14-y + z = -14 to find yy.
    −y+(−6)=−14-y + (-6) = -14
    −y−6=−14-y - 6 = -14
    −y=−14+6-y = -14 + 6
    y=8y = 8

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