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8 March, 20:52

X-6y + 4z=-12

x+y-4z=12

2x + 2y + 5z = -15

Systems of equations with three variables and three equations

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Answers (1)
  1. 8 March, 23:47
    0
    x = 0, y = 0, z = - 3

    Step-by-step explanation:

    Solve the following system:

    {x - 6 y + 4 z = - 12 | (equation 1)

    x + y - 4 z = 12 | (equation 2)

    2 x + 2 y + 5 z = - 15 | (equation 3)

    Swap equation 1 with equation 3:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    x + y - 4 z = 12 | (equation 2)

    x - 6 y + 4 z = - 12 | (equation 3)

    Subtract 1/2 * (equation 1) from equation 2:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x+0 y - (13 z) / 2 = 39/2 | (equation 2)

    x - 6 y + 4 z = - 12 | (equation 3)

    Multiply equation 2 by 2/13:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x+0 y - z = 3 | (equation 2)

    x - 6 y + 4 z = - 12 | (equation 3)

    Subtract 1/2 * (equation 1) from equation 3:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x+0 y - z = 3 | (equation 2)

    0 x - 7 y + (3 z) / 2 = - 9/2 | (equation 3)

    Multiply equation 3 by 2:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x+0 y - z = 3 | (equation 2)

    0 x - 14 y + 3 z = - 9 | (equation 3)

    Swap equation 2 with equation 3:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x - 14 y + 3 z = - 9 | (equation 2)

    0 x+0 y - z = 3 | (equation 3)

    Multiply equation 3 by - 1:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x - 14 y + 3 z = - 9 | (equation 2)

    0 x+0 y+z = - 3 | (equation 3)

    Subtract 3 * (equation 3) from equation 2:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x - 14 y+0 z = 0 | (equation 2)

    0 x+0 y+z = - 3 | (equation 3)

    Divide equation 2 by - 14:

    {2 x + 2 y + 5 z = - 15 | (equation 1)

    0 x+y+0 z = 0 | (equation 2)

    0 x+0 y+z = - 3 | (equation 3)

    Subtract 2 * (equation 2) from equation 1:

    {2 x + 0 y+5 z = - 15 | (equation 1)

    0 x+y+0 z = 0 | (equation 2)

    0 x+0 y+z = - 3 | (equation 3)

    Subtract 5 * (equation 3) from equation 1:

    {2 x+0 y+0 z = 0 | (equation 1)

    0 x+y+0 z = 0 | (equation 2)

    0 x+0 y+z = - 3 | (equation 3)

    Divide equation 1 by 2:

    {x+0 y+0 z = 0 | (equation 1)

    0 x+y+0 z = 0 | (equation 2)

    0 x+0 y+z = - 3 | (equation 3)

    Collect results:

    Answer: {x = 0, y = 0, z = - 3
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