Solve {l}{x+y+z=6}{2x+y-z=1}{3x+2y+z=11}

$$\left. \begin{array}{l}{ x + y + z = 6 }\\{ 2 x + y - z = 1 }\\{ 3 x + 2 y + z = 11 }\end{array} \right.$$

Solve for x, y, z

$$x=3$$

$$y=-1$$

$$z=4$$

Short Steps Using Substitution

Solve x+y+z=6 for x.

$$x=-y-z+6$$

Substitute -y-z+6 for x in the second and third equation.

$$2\left(-y-z+6\right)+y-z=1 3\left(-y-z+6\right)+2y+z=11$$

Solve these equations for y and z respectively.

$$y=-3z+11 z=-\frac{1}{2}y+\frac{7}{2}$$

Substitute -3z+11 for y in the equation z=-\frac{1}{2}y+\frac{7}{2}.

$$z=-\frac{1}{2}\left(-3z+11\right)+\frac{7}{2}$$

Solve z=-\frac{1}{2}\left(-3z+11\right)+\frac{7}{2} for z.

$$z=4$$

Substitute 4 for z in the equation y=-3z+11.

$$y=-3\times 4+11$$

Calculate y from y=-3\times 4+11.

$$y=-1$$

Substitute -1 for y and 4 for z in the equation x=-y-z+6.

$$x=-\left(-1\right)-4+6$$

Calculate x from x=-\left(-1\right)-4+6.

$$x=3$$

The system is now solved.

$$x=3 y=-1 z=4$$