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

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

Solve for x, y, z

$$x=1$$

$$y=-1$$

$$z=-1$$

Short Steps Using Substitution

Solve x+3y+2z=-4 for x.

$$x=-3y-2z-4$$

Substitute -3y-2z-4 for x in the second and third equation.

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

Solve these equations for y and z respectively.

$$y=-1 z=-2-y$$

Substitute -1 for y in the equation z=-2-y.

$$z=-2-\left(-1\right)$$

Calculate z from z=-2-\left(-1\right).

$$z=-1$$

Substitute -1 for y and -1 for z in the equation x=-3y-2z-4.

$$x=-3\left(-1\right)-2\left(-1\right)-4$$

Calculate x from x=-3\left(-1\right)-2\left(-1\right)-4.

$$x=1$$

The system is now solved.

$$x=1 y=-1 z=-1$$