Solve {l}{-2x+y-2z=4}{2x+y=1}{-x+3y-2z=6}

$$\left. \begin{array} { l } { - 2 x + y - 2 z = 4 } \\ { 2 x + y = 1 } \\ { - x + 3 y - 2 z = 6 } \end{array} \right.$$

Solve for x, y, z

$$x=0$$

$$y=1$$

$$z = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Short Steps Using Substitution

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

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

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

$$2x+4+2x+2z=1 -x+3\left(4+2x+2z\right)-2z=6$$

Solve these equations for x and z respectively.

$$x=-\frac{3}{4}-\frac{1}{2}z z=-\frac{5}{4}x-\frac{3}{2}$$

Substitute -\frac{3}{4}-\frac{1}{2}z for x in the equation z=-\frac{5}{4}x-\frac{3}{2}.

$$z=-\frac{5}{4}\left(-\frac{3}{4}-\frac{1}{2}z\right)-\frac{3}{2}$$

Solve z=-\frac{5}{4}\left(-\frac{3}{4}-\frac{1}{2}z\right)-\frac{3}{2} for z.

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

Substitute -\frac{3}{2} for z in the equation x=-\frac{3}{4}-\frac{1}{2}z.

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

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

$$x=0$$

Substitute 0 for x and -\frac{3}{2} for z in the equation y=4+2x+2z.

$$y=4+2\times 0+2\left(-\frac{3}{2}\right)$$

Calculate y from y=4+2\times 0+2\left(-\frac{3}{2}\right).

$$y=1$$

The system is now solved.

$$x=0 y=1 z=-\frac{3}{2}$$