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-3
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avatar+76 

 

Solve the system of equations
\( \begin{align*} 5x+3z & = 1, \\ -x + y + z & = 0, \\ 3y + 2z &= 5. \end{align*}\)

 Jun 22, 2019
 #1
avatar+8336 
+1

\(\begin{pmatrix} 5&0&3|1\\ -1&1&1|0\\ 0&3&2|5\\ \end{pmatrix}\\ \sim \begin{pmatrix} 5&0&3&|1\\ 0&1&\dfrac{8}5&|\dfrac{1}{5}\\ 0&3&2&|5\\ \end{pmatrix}\\ \sim \begin{pmatrix} 5&0&3&|1\\ 0&1&\dfrac{8}5&|\dfrac{1}{5}\\ 0&0&-\dfrac{14}{5}&|\dfrac{22}{5}\\ \end{pmatrix}\\ -\dfrac{14}{5}z = \dfrac{22}5\\ \boxed{z = \dfrac{-11}{7}}\\ y + \dfrac{8}{5}z = \dfrac{1}{5}\\ \boxed{y = \dfrac{1}{5} + \dfrac{8}{5}\cdot \dfrac{11}{7} = \dfrac{19}{7}}\\ 5x + 3z = 1\\ \boxed{x = \dfrac{1-3(\dfrac{-11}{7})}{5} = \dfrac{8}{7}}\)

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 Jun 22, 2019
 #2
avatar+76 
-3

OH GOD THATS A MESS...

 

ok ill solve from there, thx!

CuteDramione  Jun 22, 2019
 #3
avatar+8336 
0

Nope, it is already solved.

\((x,y,z) = \left(\dfrac{8}{7},\dfrac{19}{7}, \dfrac{-11}{7}\right)\).

See the parts I have encircled with a box.

MaxWong  Jun 22, 2019
 #4
avatar+76 
-5

OHHHHH :( i didn't see

CuteDramione  Jun 22, 2019

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