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If you were to use the substitution method to solve the following system, choose the new system of equations that would result if x was isolated in the second equation. 3x – 5y + z = 12 x + 6y – z = –4 2x – 5y – 2z = –8
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A: 2x – 11y = 16 4x + 7y = –16 B: 4x + y = 8 –17y = 0 C: –23y + 4z = 24 –17y = 0 D: 13y – 2z = 24 7y – 4z = 0
solve for x in the second equation in terms of y and z; and plug that "value" into the x spots of the other 2
if i see it right its a toss up of the last 2 options
3z + z = 4z so id gander to say its the third one
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