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PLEASE SOMEONE HELP ME. if you were to use substitution method to solve the following system, choose the new system of equations that would result if x was isolated in the second equation. 2x+5y-2z=-1 x+3y+z=-2 3x-2y+3x=-17 A. -y-4x=3 -11y=-11 B.11y=3 7y+6x=-11 C.4x+11y=-5 -11=-11 D.-y=-5 6x+7y=-11

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so x from second equation will be x=-2-3y-z 2x+5y-2z=-1 3x-2y+3z=-17 hope that this is hence right,correct with 3z not with 3x before = so when x=-2-3y-z will get 2(-2-3y-z)+5y-2z=-1 3(-2-3y-z)-2y+3z=-17 -4-6y-2z+5y-2z=-1 -6-9y-3z-2y+3z=-17 -y-4z=3 -11y=-11 so from this result that y=1 so than -1-4z=3 so -4z=4 so z=4/(-4)=-1 so z=-1 so than x=-2-3(1)-(-1)=-2-3+1=-5+1=-4 so x=-4 x=-4 y=1 z=-1
just choose the correct answer choice A,B,C OR D
A but there need to be -4z

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Other answers:

so if you izolated x
robtobey how do you say ?
i think i typed it wrong.. let me see !
2x-5y-2z=-1 x+3y+z=-2 3x-2y+3x=-17
Correction 3x-2y+3z=17
equale -17
yes, my bad
A Mathematica solution with comments is attached.
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Thank you for the medal.
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and medal for my work ???
thank yo
thank you and good luck

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