Solve {l}{2x+y+z=2}{x+2y+z=4}{x+y+2z=6}

$$\left\{ \begin{array} { l } { 2 x + y + z = 2 } \\ { x + 2 y + z = 4 } \\ { x + y + 2 z = 6 } \end{array} \right.$$

Solve for x, y, z

$$x=-1$$

$$y=1$$

$$z=3$$

Short Steps Using Substitution

Solve 2x+y+z=2 for y.

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

Substitute -2x-z+2 for y in the second and third equation.

$$x+2\left(-2x-z+2\right)+z=4 x-2x-z+2+2z=6$$

Solve these equations for x and z respectively.

$$x=-\frac{1}{3}z z=x+4$$

Substitute -\frac{1}{3}z for x in the equation z=x+4.

$$z=-\frac{1}{3}z+4$$

Solve z=-\frac{1}{3}z+4 for z.

$$z=3$$

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

$$x=-\frac{1}{3}\times 3$$

Calculate x from x=-\frac{1}{3}\times 3.

$$x=-1$$

Substitute -1 for x and 3 for z in the equation y=-2x-z+2.

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

Calculate y from y=-2\left(-1\right)-3+2.

$$y=1$$

The system is now solved.

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