Question

Solve for (x, y, z), if there is a solution to the given system of linear equations:

x - 3y - 2z = -3

3x + 2y - z = 12

-x - y + 4z = 3




(- 4, 1, 0)


No solution


( 4, 1, 2)


(3, - 4, 2)

Answer 1

(4, 1, 2)

Explanation:

`\left\{\begin{array}{ccc}x-3y-2z=-3&(1)\\3x+2y-z=12&(2)\\-x-y+4z=3&(3)\end{array}\right\\\\\text{add both sides of the equations (1) and (3):}\\\\\underline{+\left\{\begin{array}{ccc}x-3y-2z=-3\\-x-y+4z=3\end{array}\right}\\.\qquad-4y+2z=0\qquad\text{add 4y to both sides}\\.\qquad2z=4y\qquad\text{divide both sides by 2}\\.\qquad\boxed{z=2y}\\\\\text{Put it to (2)}:\\\\3x+2y-2y=12\\3x=12\qquad\text{divide both sides by 3}\\\boxed{x=4}\\\\\text{Put the value of x to (1) and (3):}`

`\left\{\begin{array}{ccc}4-3y-2z=-3&\text{subtract 4 from both sides}\\-4-y+4z=3&\text{add 4 to both sides}\end{array}\right\\\left\{\begin{array}{ccc}-3y-2z=-7&\text{multiply both sides by 2}\\-y+4z=7\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}-6y-4z=-14\\-y+4z=7\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad-7y=-7\qquad\text{divide both sides by (-7)}\\.\qquad\boxed{y=1}\\\\\text{Put the value of y to the second equation:}`

`-1+4z=7\qquad\text{add 1 to both sides}\\4z=8\qquad\text{divide both sides by 4}\\\boxed{z=2}`