Solve {r}{x+y+z=6}{2x-y+z=3}{3x-z=0}

$$\left\{ \begin{array} { r } { x + y + z = 6 } \\ { 2 x - y + z = 3 } \\ { 3 x - z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=2$$

$$z=3$$

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.

$$2\left(-y-z+6\right)-y+z=3 3\left(-y-z+6\right)-z=0$$

Solve these equations for y and z respectively.

$$y=-\frac{1}{3}z+3 z=\frac{9}{2}-\frac{3}{4}y$$

Substitute -\frac{1}{3}z+3 for y in the equation z=\frac{9}{2}-\frac{3}{4}y.

$$z=\frac{9}{2}-\frac{3}{4}\left(-\frac{1}{3}z+3\right)$$

Solve z=\frac{9}{2}-\frac{3}{4}\left(-\frac{1}{3}z+3\right) for z.

$$z=3$$

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

$$y=-\frac{1}{3}\times 3+3$$

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

$$y=2$$

Substitute 2 for y and 3 for z in the equation x=-y-z+6.

$$x=-2-3+6$$

Calculate x from x=-2-3+6.

$$x=1$$

The system is now solved.

$$x=1 y=2 z=3$$