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

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

Solve for x, y, z

$$x=1$$

$$y=1$$

$$z=1$$

Short Steps Using Substitution

Solve 2x+y+z=4 for y.

$$y=-2x-z+4$$

Substitute -2x-z+4 for y in the second and third equation.

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

Solve these equations for x and z respectively.

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

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

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

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

$$z=1$$

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

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

Calculate x from x=-\frac{1}{3}+\frac{4}{3}.

$$x=1$$

Substitute 1 for x and 1 for z in the equation y=-2x-z+4.

$$y=-2-1+4$$

Calculate y from y=-2-1+4.

$$y=1$$

The system is now solved.

$$x=1 y=1 z=1$$