Solve {l}{5x-y+z=0}{1/2x-4y+z=1}{12x-z*5-y=0}

$$\left. \begin{array} { l } { 5 x - y + z = 0 } \\ { \frac { 1 } { 2 } x - 4 y + z = 1 } \\ { 12 x - z \cdot 5 - y = 0 } \end{array} \right.$$

Solve for x, y, z

$$x=-\frac{1}{23}\approx -0.043478261$$

$$y=-\frac{37}{138}\approx -0.268115942$$

$$z=-\frac{7}{138}\approx -0.050724638$$

Short Steps Using Substitution

Solve 5x-y+z=0 for y.

$$y=5x+z$$

Substitute 5x+z for y in the second and third equation.

$$\frac{1}{2}x-4\left(5x+z\right)+z=1 12x-z\times 5-\left(5x+z\right)=0$$

Solve these equations for x and z respectively.

$$x=-\frac{2}{39}-\frac{2}{13}z z=\frac{7}{6}x$$

Substitute -\frac{2}{39}-\frac{2}{13}z for x in the equation z=\frac{7}{6}x.

$$z=\frac{7}{6}\left(-\frac{2}{39}-\frac{2}{13}z\right)$$

Solve z=\frac{7}{6}\left(-\frac{2}{39}-\frac{2}{13}z\right) for z.

$$z=-\frac{7}{138}$$

Substitute -\frac{7}{138} for z in the equation x=-\frac{2}{39}-\frac{2}{13}z.

$$x=-\frac{2}{39}-\frac{2}{13}\left(-\frac{7}{138}\right)$$

Calculate x from x=-\frac{2}{39}-\frac{2}{13}\left(-\frac{7}{138}\right).

$$x=-\frac{1}{23}$$

Substitute -\frac{1}{23} for x and -\frac{7}{138} for z in the equation y=5x+z.

$$y=5\left(-\frac{1}{23}\right)-\frac{7}{138}$$

Calculate y from y=5\left(-\frac{1}{23}\right)-\frac{7}{138}.

$$y=-\frac{37}{138}$$

The system is now solved.

$$x=-\frac{1}{23} y=-\frac{37}{138} z=-\frac{7}{138}$$