Solve {l}{4x-y+z=10}{2x+y+3z=4}{3x+2y+2z=-7}

$$\left. \begin{array}{l}{ 4 x - y + z = 10 }\\{ 2 x + y + 3 z = 4 }\\{ 3 x + 2 y + 2 z = - 7 }\end{array} \right.$$

Solve for x, y, z

$$x=-\frac{1}{5}=-0.2$$

$$y=-7$$

$$z = \frac{19}{5} = 3\frac{4}{5} = 3.8$$

Short Steps Using Substitution

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

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

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

$$2x+4x+z-10+3z=4 3x+2\left(4x+z-10\right)+2z=-7$$

Solve these equations for x and z respectively.

$$x=\frac{7}{3}-\frac{2}{3}z z=-\frac{11}{4}x+\frac{13}{4}$$

Substitute \frac{7}{3}-\frac{2}{3}z for x in the equation z=-\frac{11}{4}x+\frac{13}{4}.

$$z=-\frac{11}{4}\left(\frac{7}{3}-\frac{2}{3}z\right)+\frac{13}{4}$$

Solve z=-\frac{11}{4}\left(\frac{7}{3}-\frac{2}{3}z\right)+\frac{13}{4} for z.

$$z=\frac{19}{5}$$

Substitute \frac{19}{5} for z in the equation x=\frac{7}{3}-\frac{2}{3}z.

$$x=\frac{7}{3}-\frac{2}{3}\times \frac{19}{5}$$

Calculate x from x=\frac{7}{3}-\frac{2}{3}\times \frac{19}{5}.

$$x=-\frac{1}{5}$$

Substitute -\frac{1}{5} for x and \frac{19}{5} for z in the equation y=4x+z-10.

$$y=4\left(-\frac{1}{5}\right)+\frac{19}{5}-10$$

Calculate y from y=4\left(-\frac{1}{5}\right)+\frac{19}{5}-10.

$$y=-7$$

The system is now solved.

$$x=-\frac{1}{5} y=-7 z=\frac{19}{5}$$