Solve {r}{x+y-z=0}{3x+2y+z=4}{x-3y+4z=5}

$$\left\{ \begin{array} { r } { x + y - z = 0 } \\ { 3 x + 2 y + z = 4 } \\ { x - 3 y + 4 z = 5 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=0$$

$$z=1$$

Short Steps Using Substitution

Solve x+y-z=0 for x.

$$x=-y+z$$

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

$$3\left(-y+z\right)+2y+z=4 -y+z-3y+4z=5$$

Solve these equations for y and z respectively.

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

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

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

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

$$z=1$$

Substitute 1 for z in the equation y=-4+4z.

$$y=-4+4\times 1$$

Calculate y from y=-4+4\times 1.

$$y=0$$

Substitute 0 for y and 1 for z in the equation x=-y+z.

$$x=-0+1$$

Calculate x from x=-0+1.

$$x=1$$

The system is now solved.

$$x=1 y=0 z=1$$