Solve {l}{2x+y+z=-1}{0x+3y+2z=5}{-x+2y-z=3}

$$\left. \begin{array} { l } { 2 x + y + z = - 1 } \\ { 0 x + 3 y + 2 z = 5 } \\ { - x + 2 y - z = 3 } \end{array} \right.$$

Solve for x, y, z

$$x = -\frac{19}{13} = -1\frac{6}{13} \approx -1.461538462$$

$$y = \frac{15}{13} = 1\frac{2}{13} \approx 1.153846154$$

$$z=\frac{10}{13}\approx 0.769230769$$

Short Steps Using Substitution

Solve 2x+y+z=-1 for y.

$$y=-2x-z-1$$

Substitute -2x-z-1 for y in the second and third equation.

$$0x+3\left(-2x-z-1\right)+2z=5 -x+2\left(-2x-z-1\right)-z=3$$

Solve these equations for x and z respectively.

$$x=-\frac{1}{6}z-\frac{4}{3} z=-\frac{5}{3}-\frac{5}{3}x$$

Substitute -\frac{1}{6}z-\frac{4}{3} for x in the equation z=-\frac{5}{3}-\frac{5}{3}x.

$$z=-\frac{5}{3}-\frac{5}{3}\left(-\frac{1}{6}z-\frac{4}{3}\right)$$

Solve z=-\frac{5}{3}-\frac{5}{3}\left(-\frac{1}{6}z-\frac{4}{3}\right) for z.

$$z=\frac{10}{13}$$

Substitute \frac{10}{13} for z in the equation x=-\frac{1}{6}z-\frac{4}{3}.

$$x=-\frac{1}{6}\times \frac{10}{13}-\frac{4}{3}$$

Calculate x from x=-\frac{1}{6}\times \frac{10}{13}-\frac{4}{3}.

$$x=-\frac{19}{13}$$

Substitute -\frac{19}{13} for x and \frac{10}{13} for z in the equation y=-2x-z-1.

$$y=-2\left(-\frac{19}{13}\right)-\frac{10}{13}-1$$

Calculate y from y=-2\left(-\frac{19}{13}\right)-\frac{10}{13}-1.

$$y=\frac{15}{13}$$

The system is now solved.

$$x=-\frac{19}{13} y=\frac{15}{13} z=\frac{10}{13}$$