Solve {r}{x+y+t=0}{2x-y-2z=1}{x+z=3}

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

Solve for x, y, z

$$x=\frac{7-t}{5}$$

$$y=\frac{-4t-7}{5}$$

$$z=\frac{t+8}{5}$$

Short Steps Using Substitution

Solve x+y+t=0 for x.

$$x=-y-t$$

Substitute -y-t for x in the second and third equation.

$$2\left(-y-t\right)-y-2z=1 -y-t+z=3$$

Solve these equations for y and z respectively.

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

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

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

Solve z=-\frac{1}{3}-\frac{2}{3}z-\frac{2}{3}t+t+3 for z.

$$z=\frac{8}{5}+\frac{1}{5}t$$

Substitute \frac{8}{5}+\frac{1}{5}t for z in the equation y=-\frac{1}{3}-\frac{2}{3}z-\frac{2}{3}t.

$$y=-\frac{1}{3}-\frac{2}{3}\left(\frac{8}{5}+\frac{1}{5}t\right)-\frac{2}{3}t$$

Calculate y from y=-\frac{1}{3}-\frac{2}{3}\left(\frac{8}{5}+\frac{1}{5}t\right)-\frac{2}{3}t.

$$y=-\frac{7}{5}-\frac{4}{5}t$$

Substitute -\frac{7}{5}-\frac{4}{5}t for y in the equation x=-y-t.

$$x=-\left(-\frac{7}{5}-\frac{4}{5}t\right)-t$$

Calculate x from x=-\left(-\frac{7}{5}-\frac{4}{5}t\right)-t.

$$x=\frac{7}{5}-\frac{1}{5}t$$

The system is now solved.

$$x=\frac{7}{5}-\frac{1}{5}t y=-\frac{7}{5}-\frac{4}{5}t z=\frac{8}{5}+\frac{1}{5}t$$