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Find the ordered triplet (x,y,z) for the following system of equations:

x + 3y + 2z = 1

-3x + y + 5z = 10

-2x + 3y + z = 7

 Aug 17, 2021
 #1
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The solution is (x,y,z) = (2,3,-2).

 Aug 17, 2021
 #2
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Find the ordered triplet (x,y,z) for the system of equations.

 

Hello Guest!

 

  \(x + 3y + 2z = 1\ |\ \times 2\) 

     -3x + y + 5z = 10

      -2x + 3y + z = 7

 

  \(2x+6y+4z=2\)

\(\underline{-2x + 3y + z = 7}\)

           \(9y+5z=9\ |\ \times 10\)

 

  \(x + 3y + 2z = 1\ |\ \times 3\\ \ \\ 3x+9y+6z=3\)

\( \underline{ -3x + y + 5z = 10}\)

       \(10y+11z=13\ |\ \times 9\)

 

  \(90y+50z=90\)

  \(\underline{90y+99z=117}\)

              \(49z=27\)

                 \(z=\dfrac{27}{49}\)

 

 

 

   \(10y+11z=13\ |\ insert\ z\)

  \(10y+\dfrac{11\cdot 27}{49}=13\\ 490y=13\cdot49-11\cdot 27=340\)

              \(y=\dfrac{34}{49}\)

 

  \(x + 3y + 2z = 1\\ x=1-3y-2z\ |\ insert\ y\ and\ z\\ x=1-\frac{3\cdot 34}{49}-\frac{2\cdot 27}{49}\\ 49x=49-3\cdot 34-2\cdot27\\ 49x=-107\) 

          \(x=-\dfrac{107}{49}\)

 

  \((x,y,z)=(-\dfrac{107}{49},\dfrac{34}{49},\dfrac{27}{49})\)

laugh  !

 Aug 17, 2021
edited by asinus  Aug 17, 2021

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