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Solve the system of equations by elimination.\newline2xy+z=62x - y + z = -6\newline3x2y+2z=43x - 2y + 2z = -4\newlinex+3y2z=20-x + 3y - 2z = -20

Full solution

Q. Solve the system of equations by elimination.\newline2xy+z=62x - y + z = -6\newline3x2y+2z=43x - 2y + 2z = -4\newlinex+3y2z=20-x + 3y - 2z = -20
  1. Add equations to eliminate z: First, let's add the first and third equations to eliminate z.\newline(2xy+z)+(x+3y2z)=6+(20)(2x - y + z) + (-x + 3y - 2z) = -6 + (-20)\newline2xy+zx+3y2z=262x - y + z - x + 3y - 2z = -26\newlinex+2yz=26x + 2y - z = -26
  2. Multiply and add equations: Now, let's multiply the first equation by 22 and add it to the second equation to eliminate zz.
    (2(2xy+z))+(3x2y+2z)=2(6)+(4)(2*(2x - y + z)) + (3x - 2y + 2z) = 2*(-6) + (-4)
    4x2y+2z+3x2y+2z=1244x - 2y + 2z + 3x - 2y + 2z = -12 - 4
    7x4y+4z=167x - 4y + 4z = -16
    But wait, I made a mistake here, I should have eliminated zz, not kept it.

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