Solve {r}{-x+2y+z=3}{x-3y+z=0}{2x-y+6z=7}

$$\left. \begin{array} { r } { - x + 2 y + z = 3 } \\ { x - 3 y + z = 0 } \\ { 2 x - y + 6 z = 7 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{8}{7} = -1\frac{1}{7} \approx -1.142857143$$

$$y=\frac{1}{7}\approx 0.142857143$$

$$z = \frac{11}{7} = 1\frac{4}{7} \approx 1.571428571$$

Short Steps Using Substitution

Solve -x+2y+z=3 for x.

$$x=2y+z-3$$

Substitute 2y+z-3 for x in the second and third equation.

$$2y+z-3-3y+z=0 2\left(2y+z-3\right)-y+6z=7$$

Solve these equations for y and z respectively.

$$y=2z-3 z=-\frac{3}{8}y+\frac{13}{8}$$

Substitute 2z-3 for y in the equation z=-\frac{3}{8}y+\frac{13}{8}.

$$z=-\frac{3}{8}\left(2z-3\right)+\frac{13}{8}$$

Solve z=-\frac{3}{8}\left(2z-3\right)+\frac{13}{8} for z.

$$z=\frac{11}{7}$$

Substitute \frac{11}{7} for z in the equation y=2z-3.

$$y=2\times \frac{11}{7}-3$$

Calculate y from y=2\times \frac{11}{7}-3.

$$y=\frac{1}{7}$$

Substitute \frac{1}{7} for y and \frac{11}{7} for z in the equation x=2y+z-3.

$$x=2\times \frac{1}{7}+\frac{11}{7}-3$$

Calculate x from x=2\times \frac{1}{7}+\frac{11}{7}-3.

$$x=-\frac{8}{7}$$

The system is now solved.

$$x=-\frac{8}{7} y=\frac{1}{7} z=\frac{11}{7}$$