Solve {l}{x+y+z=3}{2x+y-z=6}{x+2y+3z=2}

$$\left. \begin{array} { l } { x + y + z = 3 } \\ { 2 x + y - z = 6 } \\ { x + 2 y + 3 z = 2 } \end{array} \right.$$

Solve for x, y, z

$$x=5$$

$$y=-3$$

$$z=1$$

Short Steps Using Substitution

Solve x+y+z=3 for x.

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

Substitute -y-z+3 for x in the second and third equation.

$$2\left(-y-z+3\right)+y-z=6 -y-z+3+2y+3z=2$$

Solve these equations for y and z respectively.

$$y=-3z z=-\frac{1}{2}y-\frac{1}{2}$$

Substitute -3z for y in the equation z=-\frac{1}{2}y-\frac{1}{2}.

$$z=-\frac{1}{2}\left(-3\right)z-\frac{1}{2}$$

Solve z=-\frac{1}{2}\left(-3\right)z-\frac{1}{2} for z.

$$z=1$$

Substitute 1 for z in the equation y=-3z.

$$y=-3$$

Substitute -3 for y and 1 for z in the equation x=-y-z+3.

$$x=-\left(-3\right)-1+3$$

Calculate x from x=-\left(-3\right)-1+3.

$$x=5$$

The system is now solved.

$$x=5 y=-3 z=1$$