Question #197897

Solve the following system of linear equations: x + y + z = 2 (1) 6x − 4y + 5z = 31 (2) 5x + 2y + 2z = 13 (3) The solution of the system is the ordered triple: [1] (3, − 2, 1). [2] (3, 2, 1). [3] (0, 2, 0). [4] (1, 2, 1).


1
Expert's answer
2021-05-27T07:31:26-0400

x+y+z=2 ......[1]6x4y+5z=31 .....[2]5x+2y+2z=13  .....[3]x + y + z = 2\ ......[1] \\ 6x - 4y + 5z = 31\ .....[2] \\ 5x + 2y + 2z = 13\ \ .....[3]


By Substitution method:


Solve equation [1] for the variable  z 

z=xy+2z = -x - y + 2


Plug this in for variable  z  in equation [2]

6x4y+5(xy+2)=31x9y=216x - 4y + 5\cdot(-x -y +2) = 31\\ \Rightarrow x - 9y = 21


Plug this in for variable  z  in equation [3]

5x+2y+2(xy+2)=133x=95x + 2y + 2\cdot(-x -y +2) = 13\\ \Rightarrow 3x = 9


 Solve equation [3] for the variable  x 

3x=9x=33x = 9 \\ \Rightarrow \boxed {x = 3}


 Plug this in for variable  x  in equation [2]

(3)9y=219y=18(3) - 9y = 21\\ \Rightarrow - 9y = 18


Solve equation [2] for the variable  y 

9y=18y=29y = - 18 \\ \Rightarrow \boxed{ y = - 2 }


 By now we know this much :

x=3y=2z=xy+2x = 3\\ y = -2\\ z = -x-y+2


Hence, Use the  x  and  y  values to solve for  z 

z=(3)(2)+2=1z=1z = -(3)-(-2)+2 = 1 \\\Rightarrow\boxed{z=1}


Answer: [1] (x,y,z)=(3,2,1)\boxed{(x,y,z)=(3,-2,1)}


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