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anonymous
 3 years 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)}
anonymous
 3 years 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)}

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then multiply the last one times 2 so the z's will cancel

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0Cramer'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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or you could do it that way. both take about as long to me...

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0I see, so we're just ignoring problem statements. :(

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.02x + 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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