Solve {l}{x+y-3z+t=2c}{3x-y+z-t=2a}{-x+3y-z+t=2b}

$$\left\{ \begin{array} { l } { x + y - 3 z + t = 2 c } \\ { 3 x - y + z - t = 2 a } \\ { - x + 3 y - z + t = 2 b } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{t+b+c+4a}{6}$$

$$y=\frac{-t+2a+5b-c}{6}$$

$$z=\frac{t+a+b-2c}{3}$$

Short Steps Using Substitution

Solve x+y-3z+t=2c for x.

$$x=-y+3z-t+2c$$

Substitute -y+3z-t+2c for x in the second and third equation.

$$3\left(-y+3z-t+2c\right)-y+z-t=2a -\left(-y+3z-t+2c\right)+3y-z+t=2b$$

Solve these equations for y and z respectively.

$$y=-t+\frac{5}{2}z-\frac{1}{2}a+\frac{3}{2}c z=y-\frac{1}{2}b-\frac{1}{2}c+\frac{1}{2}t$$

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

$$z=-t+\frac{5}{2}z-\frac{1}{2}a+\frac{3}{2}c-\frac{1}{2}b-\frac{1}{2}c+\frac{1}{2}t$$

Solve z=-t+\frac{5}{2}z-\frac{1}{2}a+\frac{3}{2}c-\frac{1}{2}b-\frac{1}{2}c+\frac{1}{2}t for z.

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

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

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

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

$$y=-\frac{1}{6}t-\frac{1}{6}c+\frac{1}{3}a+\frac{5}{6}b$$

Substitute -\frac{1}{6}t-\frac{1}{6}c+\frac{1}{3}a+\frac{5}{6}b for y and \frac{1}{3}t-\frac{2}{3}c+\frac{1}{3}a+\frac{1}{3}b for z in the equation x=-y+3z-t+2c.

$$x=-\left(-\frac{1}{6}t-\frac{1}{6}c+\frac{1}{3}a+\frac{5}{6}b\right)+3\left(\frac{1}{3}t-\frac{2}{3}c+\frac{1}{3}a+\frac{1}{3}b\right)-t+2c$$

Calculate x from x=-\left(-\frac{1}{6}t-\frac{1}{6}c+\frac{1}{3}a+\frac{5}{6}b\right)+3\left(\frac{1}{3}t-\frac{2}{3}c+\frac{1}{3}a+\frac{1}{3}b\right)-t+2c.

$$x=\frac{1}{6}t+\frac{1}{6}c+\frac{2}{3}a+\frac{1}{6}b$$

The system is now solved.

$$x=\frac{1}{6}t+\frac{1}{6}c+\frac{2}{3}a+\frac{1}{6}b y=-\frac{1}{6}t-\frac{1}{6}c+\frac{1}{3}a+\frac{5}{6}b z=\frac{1}{3}t-\frac{2}{3}c+\frac{1}{3}a+\frac{1}{3}b$$