Solve {l}{x+y-z-2r=0}{x+7y-5z-2r=0}{2x-y+2z+t=0}

$$\left\{ \begin{array} { l } { x + y - z - 2 r = 0 } \\ { x + 7 y - 5 z - 2 r = 0 } \\ { 2 x - y + 2 z + t = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{4r}{3}-\frac{t}{6}$$

$$y=\frac{-4r-t}{3}$$

$$z=-\frac{t}{2}-2r$$

Short Steps Using Substitution

Solve x+y-z-2r=0 for x.

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

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

$$-y+z+2r+7y-5z-2r=0 2\left(-y+z+2r\right)-y+2z+t=0$$

Solve these equations for y and z respectively.

$$y=\frac{2}{3}z z=\frac{3}{4}y-r-\frac{1}{4}t$$

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

$$z=\frac{3}{4}\times \frac{2}{3}z-r-\frac{1}{4}t$$

Solve z=\frac{3}{4}\times \frac{2}{3}z-r-\frac{1}{4}t for z.

$$z=-2r-\frac{1}{2}t$$

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

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

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

$$y=-\frac{4}{3}r-\frac{1}{3}t$$

Substitute -\frac{4}{3}r-\frac{1}{3}t for y and -2r-\frac{1}{2}t for z in the equation x=-y+z+2r.

$$x=-\left(-\frac{4}{3}r-\frac{1}{3}t\right)-2r-\frac{1}{2}t+2r$$

Calculate x from x=-\left(-\frac{4}{3}r-\frac{1}{3}t\right)-2r-\frac{1}{2}t+2r.

$$x=\frac{4}{3}r-\frac{1}{6}t$$

The system is now solved.

$$x=\frac{4}{3}r-\frac{1}{6}t y=-\frac{4}{3}r-\frac{1}{3}t z=-2r-\frac{1}{2}t$$