Question

Solve the system algebraically.

x − y + 2z = -3

2x + 2y − z = -3

3x + 2y = 8

No solution

(-42, 67, 53)

Infinite solutions

(43, -57, 63)

Answer 1

1x - 1y + 2z = -3 → 1x - 1y + 2z = -3 → 2x - 2y + 4z = -6
2x + 2y - 1z = -3 → 2x + 2y - 1z = -3 → 2x + 2y - 1z = -3
3x + 2y + 0z = 8 4x + 3z = -9

1x - 1y + 2z = -3
2x + 2y - 1z = -3 → 2x + 2y - 1z = -3 → -2x - 2y + 1z = 3
3x + 2y + 0z = 8 → 3x + 2y + 0z = 8 → 3x + 2y + 0z = 8
x + z = 11

4x + 3z = -9 → 4x + 3z = -9
1x + 1z = 11 → 4x + 4z = 44
-z = -53
-1 -1
z = 53
x + z = 11
x + 53 = 11
- 53 - 53
x = -42

3x + 2y = 8
3(-42) + 2y = 8
-126 + 2y = 8
+ 126 + 126
2y = 134
2 2
y = 67
(x, y, z) = (-42, 67, 53)

The answer is B.