If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and third equations, then the second and third equations. x – y + 2z = –2 2x + 2y + z = 7 3x + 3y – z = 3

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add 2nd and 3rd equation, what u get ?

x - y + 2z = - 2 2x + 2y + z = 7 3x + 3y - z = 3 -------------- Step 1: add 1st and 3rd equation multiplying if necessary to eliminate z x - y + 2z = -2 3x + 3y -z = 3 --->(2)3x + 3y - z = 3 --------------- x - y + 2z = -2 6x + 6y - 2z = 6 (result of multiplying 2nd equation by 2) -------------- 7x + 5y = 4 ====> new equation resulting from 1st and 3rd equation Now for the 2nd and 3rd equation..... 2x + 2y + z = 7 3x + 3y - z = 3 -------------no multiplying is needed because the z's cancel out 5x + 5y = 10 ===> new equation resulting from 2nd and 3rd equation easy enough :)

thanks That helped a lot.

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