Question

Use the elimination method to solve the following system of equations.

2x – y + z = –3
2x + 2y + 3z = 2
3x – 3y – z = –4

Answer 1

Solve
3 x-3 y-z = -4 (equation 1)
2 x+2 y+3 z = 2 (equation 2)
2 x-y+z = -3 (equation 3)

Subtract 2/3 x (equation 1) from equation 2:
3 x-3 y-z = -4 (equation 1)
0 x+4 y+(11 z)/3 = 14/3 (equation 2)
2 x-y+z = -3 (equation 3)

Multiply equation 2 by 3:

3 x-3 y-z = -4 (equation 1)
0 x+12 y+11 z = 14 (equation 2)
2 x-y+z = -3 (equation 3)

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

3 x-3 y-z = -4 | (equation 1)
0 x+12 y+11 z = 14 | (equation 2)
0 x+y+(5 z)/3 = -1/3 | (equation 3)

Multiply equation 3 by 3:

3 x-3 y-z = -4 | (equation 1)
0 x+12 y+11 z = 14 | (equation 2)
0 x+3 y+5 z = -1 | (equation 3)

Subtract 1/4 × (equation 2) from equation 3:
3 x-3 y-z = -4 (equation 1)
0 x+12 y+11 z = 14 (equation 2)
0 x+0 y+(9 z)/4 = -9/2 (equation 3)

Multiply equation 3 by 4/9:

3 x-3 y-z = -4 (equation 1)
0 x+12 y+11 z = 14 (equation 2)
0 x+0 y+z = -2 (equation 3)

Subtract 11 × (equation 3) from equation 2:

3 x-3 y-z = -4 (equation 1)
0 x+12 y+0 z = 36 (equation 2)
0 x+0 y+z = -2 (equation 3)

Divide equation 2 by 12:

3 x-3 y-z = -4 (equation 1)
0 x+y+0 z = 3 (equation 2)
0 x+0 y+z = -2 (equation 3)

Add 3 × (equation 2) to equation 1:

3 x+0 y-z = 5 (equation 1)
0 x+y+0 z = 3 (equation 2)
0 x+0 y+z = -2 (equation 3)

Add equation 3 to equation 1:

3 x+0 y+0 z = 3 (equation 1)
0 x+y+0 z = 3 (equation 2)
0 x+0 y+z = -2 (equation 3)

Divide equation 1 by 3:

{x+0 y+0 z = 1 (equation 1)
0 x+y+0 z = 3 (equation 2)
0 x+0 y+z = -2 (equation 3)
Answer: x = 1 , y = 3 , z = -2

Answer 2

thank you so muchhhh

Answer 3

no problem :)