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anonymous
 3 years ago
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
anonymous
 3 years ago
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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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1add 2nd and 3rd equation, what u get ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0x  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 :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks That helped a lot.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no problem...anytime :)
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