Question

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)

Answer 1

Answer: the person before me should be flagged

Explanation:

Answer 2

x = 6 , y = -4 , Z = -1

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