х — 4у — 22 y + 2z - 2x + 7y + (k² – 2)z 2 - Use the system of equations 2 for problems 9 and 10. k – 4 9. The value(s) of k such that the system has a unique solution is (are): (a) k + 2 (b) k = -1,2...

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Extracted text: х — 4у — 22 y + 2z - 2x + 7y + (k² – 2)z 2 - Use the system of equations 2 for problems 9 and 10. k – 4 9. The value(s) of k such that the system has a unique solution is (are): (a) k + 2 (b) k = -1,2 (c) k # ±2 (d) k + –2 (e) None of the above. 10. The value(s) of k such that the system has no solution is (are): (а) k %3D 2, —3 (b) k = 0 (c) k = 2 (d) k = -2 (e) None of the above.