Solve {r}{x+4y+3z=10}{4x+2y-2z=-2}{3x-y+z=11}

$$\left. \begin{array} { r } { x + 4 y + 3 z = 10 } \\ { 4 x + 2 y - 2 z = - 2 } \\ { 3 x - y + z = 11 } \end{array} \right.$$

Solve for x, y, z

$$x=2$$

$$y=-1$$

$$z=4$$

Short Steps Using Substitution

Solve x+4y+3z=10 for x.

$$x=-4y-3z+10$$

Substitute -4y-3z+10 for x in the second and third equation.

$$4\left(-4y-3z+10\right)+2y-2z=-2 3\left(-4y-3z+10\right)-y+z=11$$

Solve these equations for y and z respectively.

$$y=3-z z=-\frac{13}{8}y+\frac{19}{8}$$

Substitute 3-z for y in the equation z=-\frac{13}{8}y+\frac{19}{8}.

$$z=-\frac{13}{8}\left(3-z\right)+\frac{19}{8}$$

Solve z=-\frac{13}{8}\left(3-z\right)+\frac{19}{8} for z.

$$z=4$$

Substitute 4 for z in the equation y=3-z.

$$y=3-4$$

Calculate y from y=3-4.

$$y=-1$$

Substitute -1 for y and 4 for z in the equation x=-4y-3z+10.

$$x=-4\left(-1\right)-3\times 4+10$$

Calculate x from x=-4\left(-1\right)-3\times 4+10.

$$x=2$$

The system is now solved.

$$x=2 y=-1 z=4$$