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MathematicsMathematics
26 April, 00:16

Solve the following system of equation

2x + 3y - Z = 1

3x + y + 2Z = 12

x + 2y - 3 = - 5

A) (3, 1, 2)

B) (-3, 1, 2)

C) (3, - 1, 2)

D) (3, 1, - 2)

+2
Answers (1)
  1. 26 April, 01:46
    0
    x = 6, y = - 4, Z = - 1

    Step-by-step explanation:

    Solve the following system:

    {2 x + 3 y - Z = 1 | (equation 1)

    3 x + y + 2 Z = 12 | (equation 2)

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

    Express the system in standard form:

    {2 x + 3 y - Z = 1 | (equation 1)

    3 x + y + 2 Z = 12 | (equation 2)

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

    Swap equation 1 with equation 2:

    {3 x + y + 2 Z = 12 | (equation 1)

    2 x + 3 y - Z = 1 | (equation 2)

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

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

    {3 x + y + 2 Z = 12 | (equation 1)

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

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

    Multiply equation 2 by 3/7:

    {3 x + y + 2 Z = 12 | (equation 1)

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

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

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

    {3 x + y + 2 Z = 12 | (equation 1)

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

    0 x + (5 y) / 3 - (2 Z) / 3 = - 6 | (equation 3)

    Multiply equation 3 by 3:

    {3 x + y + 2 Z = 12 | (equation 1)

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

    0 x+5 y - 2 Z = - 18 | (equation 3)

    Swap equation 2 with equation 3:

    {3 x + y + 2 Z = 12 | (equation 1)

    0 x+5 y - 2 Z = - 18 | (equation 2)

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

    Subtract 1/5 * (equation 2) from equation 3:

    {3 x + y + 2 Z = 12 | (equation 1)

    0 x+5 y - 2 Z = - 18 | (equation 2)

    0 x+0 y - (3 Z) / 5 = 3/5 | (equation 3)

    Multiply equation 3 by 5/3:

    {3 x + y + 2 Z = 12 | (equation 1)

    0 x+5 y - 2 Z = - 18 | (equation 2)

    0 x+0 y - Z = 1 | (equation 3)

    Multiply equation 3 by - 1:

    {3 x + y + 2 Z = 12 | (equation 1)

    0 x+5 y - 2 Z = - 18 | (equation 2)

    0 x+0 y+Z = - 1 | (equation 3)

    Add 2 * (equation 3) to equation 2:

    {3 x + y + 2 Z = 12 | (equation 1)

    0 x+5 y+0 Z = - 20 | (equation 2)

    0 x+0 y+Z = - 1 | (equation 3)

    Divide equation 2 by 5:

    {3 x + y + 2 Z = 12 | (equation 1)

    0 x+y+0 Z = - 4 | (equation 2)

    0 x+0 y+Z = - 1 | (equation 3)

    Subtract equation 2 from equation 1:

    {3 x + 0 y+2 Z = 16 | (equation 1)

    0 x+y+0 Z = - 4 | (equation 2)

    0 x+0 y+Z = - 1 | (equation 3)

    Subtract 2 * (equation 3) from equation 1:

    {3 x+0 y+0 Z = 18 | (equation 1)

    0 x+y+0 Z = - 4 | (equation 2)

    0 x+0 y+Z = - 1 | (equation 3)

    Divide equation 1 by 3:

    {x+0 y+0 Z = 6 | (equation 1)

    0 x+y+0 Z = - 4 | (equation 2)

    0 x+0 y+Z = - 1 | (equation 3)

    Collect results:

    Answer: {x = 6, y = - 4, Z = - 1
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Home » Mathematics » Solve the following system of equation 2x + 3y - Z = 1 3x + y + 2Z = 12 x + 2y - 3 = - 5 A) (3, 1, 2) B) (-3, 1, 2) C) (3, - 1, 2) D) (3, 1, - 2)
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