Solve {l}{x+y+z=1}{x+z=2y}{x=z}

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

Solve for x, y, z

$$x=\frac{1}{3}\approx 0.333333333$$

$$y=\frac{1}{3}\approx 0.333333333$$

$$z=\frac{1}{3}\approx 0.333333333$$

Short Steps Using Substitution

Reorder the equations.

$$x=z x+z=2y x+y+z=1$$

Substitute z for x in the second and third equation.

$$z+z=2y z+y+z=1$$

Solve these equations for y and z respectively.

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

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

$$z=-\frac{1}{2}z+\frac{1}{2}$$

Solve z=-\frac{1}{2}z+\frac{1}{2} for z.

$$z=\frac{1}{3}$$

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

$$y=\frac{1}{3}$$

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

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

The system is now solved.

$$x=\frac{1}{3} y=\frac{1}{3} z=\frac{1}{3}$$