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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)}
 one year ago
 one year ago
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)}
 one year ago
 one year ago

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brandonlovesBest ResponseYou've already chosen the best response.0
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+8y2z=44 and then add it to the second
 one year ago

brandonlovesBest ResponseYou've already chosen the best response.0
then multiply the last one times 2 so the z's will cancel
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
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.
 one year ago

brandonlovesBest ResponseYou've already chosen the best response.0
or you could do it that way. both take about as long to me...
 one year ago

tkhunnyBest ResponseYou've already chosen the best response.0
I see, so we're just ignoring problem statements. :(
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
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 :(
 one year ago
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