Use Cramer's rule to solve the system. 2x + 4y - z = 32 x - 2y + 2z = -5 5x + y + z = 20 A. {( 1, -9, -6)} B. {( 2, 7, 6)} C. {( 9, 6, 9)} D. {( 1, 9, 6)}

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Use Cramer's rule to solve the system. 2x + 4y - z = 32 x - 2y + 2z = -5 5x + y + z = 20 A. {( 1, -9, -6)} B. {( 2, 7, 6)} C. {( 9, 6, 9)} D. {( 1, 9, 6)}

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we can start by clearing the z on the first two for starters. we do that by multiplying 2 to the first problem which makes it 4x+8y-2z=44 and then add it to the second
I got 5x+6y=39
then multiply the last one times -2 so the z's will cancel

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Cramer's Rule? Well, I suppose we could simplify it a little, first, but that is a little unusual. You have four 3x3 determinates in your future.
or you could do it that way. both take about as long to me...
I see, so we're just ignoring problem statements. :-(
2x + 4y - z = 32 x - 2y + 2z = -5 5x + y + z = 20 \[\Delta = \left| \begin{matrix}2 & 4 &-1\\ 1 & -2 & 2 \\ 5 &1 & 1\end{matrix}\right|=...\] \[\Delta _{x} = \left| \begin{matrix}32 & 4 &-1\\ -5 & -2 & 2 \\ 20 &1 & 1\end{matrix}\right|=...\] \[\Delta _{y} = \left| \begin{matrix}2 & 32 &-1\\ 1 & -5 & 2 \\ 5 & 20 & 1\end{matrix}\right|=...\] \[\Delta _{z}= \left| \begin{matrix}2 & 4 & 32\\ 1 & -2 & -5 \\ 5 &1 & 20\end{matrix}\right|=...\] \[x =\frac{\Delta _{x}}{\Delta}\]\[y =\frac{\Delta _{y}}{\Delta}\]\[z =\frac{\Delta _{z}}{\Delta}\] Haven't used Cramer's Rule for long :(

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