Solve {l}{4x+3y+2z=0}{8x+2y+3z=0}{4x+2y+39z=8}

$$\left. \begin{array} { l } { 4 x + 3 y + 2 z = 0 } \\ { 8 x + 2 y + 3 z = 0 } \\ { 4 x + 2 y + 39 z = 8 } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{10}{149}\approx -0.067114094$$

$$y=-\frac{8}{149}\approx -0.053691275$$

$$z=\frac{32}{149}\approx 0.214765101$$

Short Steps Using Substitution

Solve 4x+3y+2z=0 for x.

$$x=-\frac{3}{4}y-\frac{1}{2}z$$

Substitute -\frac{3}{4}y-\frac{1}{2}z for x in the second and third equation.

$$8\left(-\frac{3}{4}y-\frac{1}{2}z\right)+2y+3z=0 4\left(-\frac{3}{4}y-\frac{1}{2}z\right)+2y+39z=8$$

Solve these equations for y and z respectively.

$$y=-\frac{1}{4}z z=\frac{8}{37}+\frac{1}{37}y$$

Substitute -\frac{1}{4}z for y in the equation z=\frac{8}{37}+\frac{1}{37}y.

$$z=\frac{8}{37}+\frac{1}{37}\left(-\frac{1}{4}\right)z$$

Solve z=\frac{8}{37}+\frac{1}{37}\left(-\frac{1}{4}\right)z for z.

$$z=\frac{32}{149}$$

Substitute \frac{32}{149} for z in the equation y=-\frac{1}{4}z.

$$y=-\frac{1}{4}\times \frac{32}{149}$$

Calculate y from y=-\frac{1}{4}\times \frac{32}{149}.

$$y=-\frac{8}{149}$$

Substitute -\frac{8}{149} for y and \frac{32}{149} for z in the equation x=-\frac{3}{4}y-\frac{1}{2}z.

$$x=-\frac{3}{4}\left(-\frac{8}{149}\right)-\frac{1}{2}\times \frac{32}{149}$$

Calculate x from x=-\frac{3}{4}\left(-\frac{8}{149}\right)-\frac{1}{2}\times \frac{32}{149}.

$$x=-\frac{10}{149}$$

The system is now solved.

$$x=-\frac{10}{149} y=-\frac{8}{149} z=\frac{32}{149}$$