Solve {l}{0=x+2y+z}{0=2x+y}{1=4x+3y+2z}

$$\left. \begin{array} { l } { 0 = x + 2 y + z } \\ { 0 = 2 x + y } \\ { 1 = 4 x + 3 y + 2 z } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{1}{4}=0.25$$

$$y=-\frac{1}{2}=-0.5$$

$$z=\frac{3}{4}=0.75$$

Short Steps Using Substitution

Solve 0=x+2y+z for x.

$$x=-2y-z$$

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

$$0=2\left(-2y-z\right)+y 1=4\left(-2y-z\right)+3y+2z$$

Solve these equations for y and z respectively.

$$y=-\frac{2}{3}z z=-\frac{5}{2}y-\frac{1}{2}$$

Substitute -\frac{2}{3}z for y in the equation z=-\frac{5}{2}y-\frac{1}{2}.

$$z=-\frac{5}{2}\left(-\frac{2}{3}\right)z-\frac{1}{2}$$

Solve z=-\frac{5}{2}\left(-\frac{2}{3}\right)z-\frac{1}{2} for z.

$$z=\frac{3}{4}$$

Substitute \frac{3}{4} for z in the equation y=-\frac{2}{3}z.

$$y=-\frac{2}{3}\times \frac{3}{4}$$

Calculate y from y=-\frac{2}{3}\times \frac{3}{4}.

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

Substitute -\frac{1}{2} for y and \frac{3}{4} for z in the equation x=-2y-z.

$$x=-2\left(-\frac{1}{2}\right)-\frac{3}{4}$$

Calculate x from x=-2\left(-\frac{1}{2}\right)-\frac{3}{4}.

$$x=\frac{1}{4}$$

The system is now solved.

$$x=\frac{1}{4} y=-\frac{1}{2} z=\frac{3}{4}$$