Solve the system by elimination

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{-2x + 2y + 3z = 0

{-2x - y + z = -3

{2x + 3y + 3z = 5

Oct 6th, 2015

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first add the first and third equation

-2x + 2y + 3z = 0   +

2x + 3y + 3z = 5


so, 0 + 5 y + 6z =5.................(1)


now add the second and third equation

-2x - y + z = -3

2x + 3y + 3z = 5

so, 0 + 2y +4z = 2........................(2)


0 + 2y +4z = 2  multiply by 3

     6y +12y = 6.....................(3)


0 + 5 y + 6z =5.................(1)  multiply by 2

   10 y + 12y =10..............(4)


now subtract (3) from (4)

   10 y + 12y =10   - 

      6y +12y = 6


so,  4 y + 0 = 4

       4y = 4

         y =1


now substitute y =1 in  5 y + 6z =5.................(1)

                                     5*1 + 6z =5

                                              6z =5-5

                                               z = 0


-2x + 2y + 3z = 0

substitue value of y=1 and z=0

-2x + 2*1 + 3*0 = 0

-2x    =  -2

  x =1



ANSWER is

x =1

y =1

z = 0



hope you understood..please message if you have any doubts...thank you
Oct 6th, 2015

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