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What is the solution to the system of equations?

{ 2x+y-z=-3, 5x-2y+2z=24, 3x-z=5}?

 

Options: (4,1,8) (-1,0,-8) (2,-6,1) or (-5,4,-3) Sorry, I'm just really stuck on this one

Guest Sep 12, 2017
 #1
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I figured it out, nvm. It is actually choice 3. Lol sorry

Guest Sep 12, 2017
 #2
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What is the solution to the system of equations?

{ 2x+y-z=-3, 5x-2y+2z=24, 3x-z=5}?

 

Options: (4,1,8) (-1,0,-8) (2,-6,1) or (-5,4,-3)

 

\(\small{ \begin{array}{|lcl|} \hline x &=& \dfrac{ \begin{vmatrix} -3&1&-1 \\ 24&-2&2 \\ 5&0&-1 \\ \end{vmatrix} }{\begin{vmatrix} 2&1&-1 \\ 5&-2&2 \\ 3&0&-1 \\ \end{vmatrix} } \\\\ &=&\dfrac{ 5\cdot [1\cdot 2- (-2) \cdot (-1) ] -1 \cdot [(-3)\cdot (-2) - 24\cdot 1 ] } {3\cdot [1\cdot 2 - (-2) \cdot (-1)] -1 \cdot [(2\cdot (-2) - 5\cdot 1] } \\\\ &=&\dfrac{ 0 -1 \cdot (-18) } {0 -1 \cdot (-9) } \\\\ &=&\dfrac{ 18 } {9} \\\\ \mathbf{x} & \mathbf{=} & \mathbf{2}\\ \hline \end{array} }\)

 

\(\small{ \begin{array}{|lcl|} \hline z &=& \dfrac{ \begin{vmatrix} 2&1&-3 \\ 5&-2&24 \\ 3&0&5 \\ \end{vmatrix} }{\begin{vmatrix} 2&1&-1 \\ 5&-2&2 \\ 3&0&-1 \\ \end{vmatrix}} \\\\ &=&\dfrac{ 3\cdot [1\cdot 24- (-2) \cdot (-3) ] +5 \cdot [2\cdot (-2) - 5\cdot 1 ] } {9 } \\\\ &=&\dfrac{ 3\cdot 18 +5\cdot (-9) } {9} \\\\ &=&\dfrac{ 9 } {9} \\\\ \mathbf{z} & \mathbf{=} & \mathbf{1}\\ \hline \end{array} }\)

 

\(\begin{array}{|lcl|} \hline y &=& -3 + z - 2x \\ &=& -3 + 1 - 2\cdot 2 \\ &=& -3 + 1 - 4 \\ \mathbf{y} & \mathbf{=} & \mathbf{ -6}\\ \hline \end{array}\)

 

Result (2,-6,1)

 

laugh

heureka  Sep 12, 2017

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