Solve {l}{x+y+z=4}{x+2y+4z=12}{2x-3y-z=4}

$$\left. \begin{array} { l } { x + y + z = 4 } \\ { x + 2 y + 4 z = 12 } \\ { 2 x - 3 y - z = 4 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=-1$$

$$z=3$$

Short Steps Using Substitution

Solve x+y+z=4 for x.

$$x=-y-z+4$$

Substitute -y-z+4 for x in the second and third equation.

$$-y-z+4+2y+4z=12 2\left(-y-z+4\right)-3y-z=4$$

Solve these equations for y and z respectively.

$$y=-3z+8 z=-\frac{5}{3}y+\frac{4}{3}$$

Substitute -3z+8 for y in the equation z=-\frac{5}{3}y+\frac{4}{3}.

$$z=-\frac{5}{3}\left(-3z+8\right)+\frac{4}{3}$$

Solve z=-\frac{5}{3}\left(-3z+8\right)+\frac{4}{3} for z.

$$z=3$$

Substitute 3 for z in the equation y=-3z+8.

$$y=-3\times 3+8$$

Calculate y from y=-3\times 3+8.

$$y=-1$$

Substitute -1 for y and 3 for z in the equation x=-y-z+4.

$$x=-\left(-1\right)-3+4$$

Calculate x from x=-\left(-1\right)-3+4.

$$x=2$$

The system is now solved.

$$x=2 y=-1 z=3$$