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Solve the system of equations by substitution.\newline3x2y+2z=143x - 2y + 2z = -14\newlinexy3z=16x - y - 3z = -16\newlinex=6x = -6

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Q. Solve the system of equations by substitution.\newline3x2y+2z=143x - 2y + 2z = -14\newlinexy3z=16x - y - 3z = -16\newlinex=6x = -6
  1. Substitute x=6x = -6: Substitute x=6x = -6 into the second equation, xy3z=16x - y - 3z = -16.\newline6y3z=16-6 - y - 3z = -16.\newlineNow, solve for y.\newliney3z=16+6-y - 3z = -16 + 6.\newliney3z=10-y - 3z = -10.\newliney+3z=10y + 3z = 10.
  2. Substitute x=6x = -6: Substitute x=6x = -6 into the first equation, 3x2y+2z=143x - 2y + 2z = -14.
    3(6)2y+2z=143(-6) - 2y + 2z = -14.
    182y+2z=14-18 - 2y + 2z = -14.
    Now, solve for yy.
    2y+2z=14+18-2y + 2z = -14 + 18.
    2y+2z=4-2y + 2z = 4.
    Divide by 2-2.
    yz=2y - z = -2.
    x=6x = -600.
  3. Solve for z: We have two expressions for y: y+3z=10y + 3z = 10 and y=z2y = z - 2. Set them equal to each other to solve for z. z2+3z=10z - 2 + 3z = 10. Combine like terms. 4z2=104z - 2 = 10. Add 22 to both sides. 4z=124z = 12. Divide by 44. z=3z = 3.
  4. Find yy: Substitute z=3z = 3 into y=z2y = z - 2 to find yy.\newliney=32y = 3 - 2.\newliney=1y = 1.
  5. Final Solution: We already have x=6x = -6 and z=3z = 3, and we just found y=1y = 1. The solution is (6,1,3)(-6, 1, 3).

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