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

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

Solve for x, y, z

$$x=1$$

$$y=-1$$

$$z=2$$

Short Steps Using Substitution

Solve x+y+z=2 for x.

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

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

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

Solve these equations for y and z respectively.

$$y=-\frac{4}{3}z+\frac{5}{3} z=\frac{3}{2}-\frac{1}{2}y$$

Substitute -\frac{4}{3}z+\frac{5}{3} for y in the equation z=\frac{3}{2}-\frac{1}{2}y.

$$z=\frac{3}{2}-\frac{1}{2}\left(-\frac{4}{3}z+\frac{5}{3}\right)$$

Solve z=\frac{3}{2}-\frac{1}{2}\left(-\frac{4}{3}z+\frac{5}{3}\right) for z.

$$z=2$$

Substitute 2 for z in the equation y=-\frac{4}{3}z+\frac{5}{3}.

$$y=-\frac{4}{3}\times 2+\frac{5}{3}$$

Calculate y from y=-\frac{4}{3}\times 2+\frac{5}{3}.

$$y=-1$$

Substitute -1 for y and 2 for z in the equation x=-y-z+2.

$$x=-\left(-1\right)-2+2$$

Calculate x from x=-\left(-1\right)-2+2.

$$x=1$$

The system is now solved.

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