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Another difficult problem, I need help on. x+3y-z=-1 2y+z=0 2z=12 Answer choice: A. x = 10, y = 3, z = 6 B. x = 1, y = 3, z = 6 C. x = 14, y = -3, z = 6
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k lets start at the bottom if \(2z=12\) then what is \(z\) ?
Solve for z first. 2z = 12 --> z = 6
6 (:
@satellite73
Plug z into second equation. 2y + 6 = 0. --> 2y = -6 --> y = -3
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C. X=14, Y=-3 , Z=6...
Uh huh, then? :o
Thank's @chirag.ahir :)
lol but I kinda need to know how to solve this..
Plug both into x + 3y - z = -1. x + 3(-3) - 6 = -1 x - 9 -6 = -1 x - 15 = -1 x = 14.
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Thus, your answer will be x = 14, y = -3, and z = 6.
I hope this helped!
Thank you!! :)
No problem! I am certainly glad to help.
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