Answer to Question #197897 in Linear Algebra for Tavlene Singh

Question

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).

Answer 1

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

By Substitution method:

Solve equation [1] for the variable  z 

$z = -x - y + 2$

Plug this in for variable  z  in equation [2]

$6x - 4y + 5\cdot(-x -y +2) = 31\\ \Rightarrow x - 9y = 21$

Plug this in for variable  z  in equation [3]

$5x + 2y + 2\cdot(-x -y +2) = 13\\ \Rightarrow 3x = 9$

 Solve equation [3] for the variable  x 

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

 Plug this in for variable  x  in equation [2]

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

Solve equation [2] for the variable  y 

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

 By now we know this much :

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

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

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

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