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7 February, 10:45

Solve each system by elimination

-x-5y-5z=2

4x-5y+4z=19

x+5y-z=-20

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Answers (1)
  1. 7 February, 14:01
    0
    x = - 2, y = - 3, z = 3

    Step-by-step explanation:

    Solve the following system:

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

    4 x - 5 y + 4 z = 19 | (equation 2)

    x + 5 y - z = - 20 | (equation 3)

    Swap equation 1 with equation 2:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    -x - 5 y - 5 z = 2 | (equation 2)

    x + 5 y - z = - 20 | (equation 3)

    Add 1/4 * (equation 1) to equation 2:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - (25 y) / 4 - 4 z = 27/4 | (equation 2)

    x + 5 y - z = - 20 | (equation 3)

    Multiply equation 2 by 4:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - 25 y - 16 z = 27 | (equation 2)

    x + 5 y - z = - 20 | (equation 3)

    Subtract 1/4 * (equation 1) from equation 3:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - 25 y - 16 z = 27 | (equation 2)

    0 x + (25 y) / 4 - 2 z = - 99/4 | (equation 3)

    Multiply equation 3 by 4:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - 25 y - 16 z = 27 | (equation 2)

    0 x+25 y - 8 z = - 99 | (equation 3)

    Add equation 2 to equation 3:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - 25 y - 16 z = 27 | (equation 2)

    0 x+0 y - 24 z = - 72 | (equation 3)

    Divide equation 3 by - 24:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - 25 y - 16 z = 27 | (equation 2)

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

    Add 16 * (equation 3) to equation 2:

    {4 x - 5 y + 4 z = 19 | (equation 1)

    0 x - 25 y+0 z = 75 | (equation 2)

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

    Divide equation 2 by - 25:

    {4 x - 5 y + 4 z = 19 | (equation 1)

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

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

    Add 5 * (equation 2) to equation 1:

    {4 x + 0 y+4 z = 4 | (equation 1)

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

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

    Subtract 4 * (equation 3) from equation 1:

    {4 x+0 y+0 z = - 8 | (equation 1)

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

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

    Divide equation 1 by 4:

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

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

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

    Collect results:

    Answer: {x = - 2, y = - 3, z = 3
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