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27 June, 21:58

X+y+z=12

6x-2y+z=16

3x+4y+2z=28

What does x, y, and z equal?

+3
Answers (1)
  1. 27 June, 22:06
    0
    x = 20/13, y = 16/13, z = 120/13

    Step-by-step explanation:

    Solve the following system:

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

    6 x - 2 y + z = 16 | (equation 2)

    3 x + 4 y + 2 z = 28 | (equation 3)

    Swap equation 1 with equation 2:

    {6 x - 2 y + z = 16 | (equation 1)

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

    3 x + 4 y + 2 z = 28 | (equation 3)

    Subtract 1/6 * (equation 1) from equation 2:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x + (4 y) / 3 + (5 z) / 6 = 28/3 | (equation 2)

    3 x + 4 y + 2 z = 28 | (equation 3)

    Multiply equation 2 by 6:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+8 y + 5 z = 56 | (equation 2)

    3 x + 4 y + 2 z = 28 | (equation 3)

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

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+8 y + 5 z = 56 | (equation 2)

    0 x+5 y + (3 z) / 2 = 20 | (equation 3)

    Multiply equation 3 by 2:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+8 y + 5 z = 56 | (equation 2)

    0 x+10 y + 3 z = 40 | (equation 3)

    Swap equation 2 with equation 3:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+10 y + 3 z = 40 | (equation 2)

    0 x+8 y + 5 z = 56 | (equation 3)

    Subtract 4/5 * (equation 2) from equation 3:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+10 y + 3 z = 40 | (equation 2)

    0 x+0 y + (13 z) / 5 = 24 | (equation 3)

    Multiply equation 3 by 5:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+10 y + 3 z = 40 | (equation 2)

    0 x+0 y+13 z = 120 | (equation 3)

    Divide equation 3 by 13:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+10 y + 3 z = 40 | (equation 2)

    0 x+0 y+z = 120/13 | (equation 3)

    Subtract 3 * (equation 3) from equation 2:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+10 y+0 z = 160/13 | (equation 2)

    0 x+0 y+z = 120/13 | (equation 3)

    Divide equation 2 by 10:

    {6 x - 2 y + z = 16 | (equation 1)

    0 x+y+0 z = 16/13 | (equation 2)

    0 x+0 y+z = 120/13 | (equation 3)

    Add 2 * (equation 2) to equation 1:

    {6 x + 0 y+z = 240/13 | (equation 1)

    0 x+y+0 z = 16/13 | (equation 2)

    0 x+0 y+z = 120/13 | (equation 3)

    Subtract equation 3 from equation 1:

    {6 x+0 y+0 z = 120/13 | (equation 1)

    0 x+y+0 z = 16/13 | (equation 2)

    0 x+0 y+z = 120/13 | (equation 3)

    Divide equation 1 by 6:

    {x+0 y+0 z = 20/13 | (equation 1)

    0 x+y+0 z = 16/13 | (equation 2)

    0 x+0 y+z = 120/13 | (equation 3)

    Collect results:

    Answer: {x = 20/13, y = 16/13, z = 120/13
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