Solve {l}{2x-4y+z=0}{3x-y-2z-4=0}{x=1}

$$\left\{ \begin{array} { l } { 2 x - 4 y + z = 0 } \\ { 3 x - y - 2 z - 4 = 0 } \\ { x = 1 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=\frac{1}{3}\approx 0.333333333$$

$$z=-\frac{2}{3}\approx -0.666666667$$

Short Steps Using Substitution

Reorder the equations.

$$x=1 3x-y-2z-4=0 2x-4y+z=0$$

Substitute 1 for x in the second and third equation.

$$3\times 1-y-2z-4=0 2\times 1-4y+z=0$$

Solve these equations for y and z respectively.

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

Substitute -1-2z for y in the equation z=-2+4y.

$$z=-2+4\left(-1-2z\right)$$

Solve z=-2+4\left(-1-2z\right) for z.

$$z=-\frac{2}{3}$$

Substitute -\frac{2}{3} for z in the equation y=-1-2z.

$$y=-1-2\left(-\frac{2}{3}\right)$$

Calculate y from y=-1-2\left(-\frac{2}{3}\right).

$$y=\frac{1}{3}$$

The system is now solved.

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