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thcsbabygirl1224
 one year ago
help me please
4x+y+5z=9
x4y2x+2
2x+3y2z=21
thcsbabygirl1224
 one year ago
help me please 4x+y+5z=9 x4y2x+2 2x+3y2z=21

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thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0what its a system of equations solving for 3 variables so im lost

thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0@LazyBoy @hba

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The second equation should be x4y2z =2, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0take the third one subtract the second one, what do you get? that is (3) (2) =??

thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0i am trying to get rid of the x in all 3 equations so i have to set them to 0 im lost on the work

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0YOu lost on your way, give it a try on my way, ok? Again (equation 3 )(equation 2) is??? You should work on it, I am guiding you only.

thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0@kal1921 what is the matrics

thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0ok i have now 2x4y2z=4 2x+3y2z=21

thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here, you have a system of two equations, with three unknows. We can rewrite it as below: \[\left\{ \begin{gathered} 2x + 4y = 4  2z \hfill \\ 2x + 3y = 21 + 2z \hfill \\ \end{gathered} \right.\] which can, easily solved for x, and y as functions of z. Please try

thcsbabygirl1224
 one year ago
Best ResponseYou've already chosen the best response.0You get 7y4z=16
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