Solve {l}{3x+5y-4z=2}{1x-4z+3y=20}{4x-1z+2y=14}

$$\left. \begin{array} { l } { 3 x + 5 y - 4 z = 2 } \\ { 1 x - 4 z + 3 y = 20 } \\ { 4 x - 1 z + 2 y = 14 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{81}{10} = 8\frac{1}{10} = 8.1$$

$$y = -\frac{171}{10} = -17\frac{1}{10} = -17.1$$

$$z = -\frac{79}{5} = -15\frac{4}{5} = -15.8$$

Short Steps Using Substitution

Reorder the equations.

$$4x-z+2y=14 1x-4z+3y=20 3x+5y-4z=2$$

Solve 4x-z+2y=14 for z.

$$z=-14+4x+2y$$

Substitute -14+4x+2y for z in the second and third equation.

$$1x-4\left(-14+4x+2y\right)+3y=20 3x+5y-4\left(-14+4x+2y\right)=2$$

Solve these equations for y and x respectively.

$$y=\frac{36}{5}-3x x=\frac{54}{13}-\frac{3}{13}y$$

Substitute \frac{36}{5}-3x for y in the equation x=\frac{54}{13}-\frac{3}{13}y.

$$x=\frac{54}{13}-\frac{3}{13}\left(\frac{36}{5}-3x\right)$$

Solve x=\frac{54}{13}-\frac{3}{13}\left(\frac{36}{5}-3x\right) for x.

$$x=\frac{81}{10}$$

Substitute \frac{81}{10} for x in the equation y=\frac{36}{5}-3x.

$$y=\frac{36}{5}-3\times \frac{81}{10}$$

Calculate y from y=\frac{36}{5}-3\times \frac{81}{10}.

$$y=-\frac{171}{10}$$

Substitute -\frac{171}{10} for y and \frac{81}{10} for x in the equation z=-14+4x+2y.

$$z=-14+4\times \frac{81}{10}+2\left(-\frac{171}{10}\right)$$

Calculate z from z=-14+4\times \frac{81}{10}+2\left(-\frac{171}{10}\right).

$$z=-\frac{79}{5}$$

The system is now solved.

$$x=\frac{81}{10} y=-\frac{171}{10} z=-\frac{79}{5}$$