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

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

Solve for x, y, z

$$x=\frac{2}{5}=0.4$$

$$y = -\frac{6}{5} = -1\frac{1}{5} = -1.2$$

$$z = \frac{11}{5} = 2\frac{1}{5} = 2.2$$

Short Steps Using Substitution

Solve x+y-z+3=0 for x.

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

Substitute -y+z-3 for x in the equation 2x-y+2-4=0.

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

Solve the second equation for y and the third equation for z.

$$y=-\frac{8}{3}+\frac{2}{3}z z=-y+1$$

Substitute -\frac{8}{3}+\frac{2}{3}z for y in the equation z=-y+1.

$$z=-\left(-\frac{8}{3}+\frac{2}{3}z\right)+1$$

Solve z=-\left(-\frac{8}{3}+\frac{2}{3}z\right)+1 for z.

$$z=\frac{11}{5}$$

Substitute \frac{11}{5} for z in the equation y=-\frac{8}{3}+\frac{2}{3}z.

$$y=-\frac{8}{3}+\frac{2}{3}\times \frac{11}{5}$$

Calculate y from y=-\frac{8}{3}+\frac{2}{3}\times \frac{11}{5}.

$$y=-\frac{6}{5}$$

Substitute -\frac{6}{5} for y and \frac{11}{5} for z in the equation x=-y+z-3.

$$x=-\left(-\frac{6}{5}\right)+\frac{11}{5}-3$$

Calculate x from x=-\left(-\frac{6}{5}\right)+\frac{11}{5}-3.

$$x=\frac{2}{5}$$

The system is now solved.

$$x=\frac{2}{5} y=-\frac{6}{5} z=\frac{11}{5}$$