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)

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