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Consider the following systems of square linear systems: x+3y-z a. 4x-y+2z = 4 =8. (2x-7y+4=-3 x+3y-z = 4 b. 4x-y+22 = 8. (2x-7y+4z =

     

Consider the following systems of square linear systems: x+3y-z a. 4x-y+2z = 4 =8. (2x-7y+4=-3 x+3y-z = 4 b. 4x-y+22 = 8. (2x-7y+4z = 0 x+3y-z = 4 C. 4x-y+2z = 8. (2x-7y+6z = 0 Which of these systems has no solution? (There is only one.) What are the solutions of the remaining two systems?

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