Solve the system of equations. 2x-3y+z=-1 3x+2y+2z=-1 x-y-3z=-4

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Solve the system of equations. 2x-3y+z=-1 3x+2y+2z=-1 x-y-3z=-4

Mathematics
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this stuff is a real pain let the machine do it
you know how?
Not a clue

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Other answers:

Which machine?
http://www.wolframalpha.com/input/?i=+2x-3y%2Bz%3D-1%2C+3x%2B2y%2B2z%3D-1%2C+x-y-3z%3D-4
Thank you!
the machine you are typing on
yw
I have another question
The graph of which quadratic equation has a vertex of (-2,1). Y=(x+1)^2-2 Y=(x-1)^2-2 Y=(x+2)^2+1 Y=(x-2)^2+1
need it by hand (for question 1), then I would suggest matrix. it is already in order for a matrix.
I got the answer to the first question. And I don't know matrix
oh.
`y=(x-h)²+k` has a vertex (h,k)
now, where would this (h,k) be (-2,1) ? In which of the options?
Y=(x-1)^2+1
y=(x-h)²+k you need vertex (-2,1) and in that form the vertex is (h,k) so h=-2, k=1 y=(x- (-2) )² + (1) y= ? (you tell me....)
Um I'm not really sure
HI!!
Hello :)
where did your \(-1\) come from? \[y=(x-\color{blue}h)^2+\color{red}k\] put \[\color{blue}h=\color{blue}{-2}\] and \[\color{red}k=\color{red}1\]
My bad I meant to put 2 not 1
final answer?
That's what I'm trying to figure out and I can't seem to get it
it is a straight forward substitution plug in the numbers as above that is all
So y=(X-2)^2+1?
no dear, the minus sign comes with the number
So y=(x+2)^2+1?
\[y=(x-(-2))^2+1\] but of course you don't want to write \(-(-2)\)
yeah that one has vertex \((-2,1)\)
Thank you! I actually understand it now!
\[\color\magenta\heartsuit\]

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