Solve {l}{2x-y+z=-2}{6x+3y-4z=8}{-2x+2y+2=8}

$$\left\{ \begin{array} { l } { 2 x - y + z = - 2 } \\ { 6 x + 3 y - 4 z = 8 } \\ { - 2 x + 2 y + 2 = 8 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{3}{13}\approx 0.230769231$$

$$y = \frac{42}{13} = 3\frac{3}{13} \approx 3.230769231$$

$$z=\frac{10}{13}\approx 0.769230769$$

Short Steps Using Substitution

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

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

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

$$6x+3\left(2x+z+2\right)-4z=8 -2x+2\left(2x+z+2\right)+2=8$$

Solve these equations for x and z respectively.

$$x=\frac{1}{6}+\frac{1}{12}z z=1-x$$

Substitute \frac{1}{6}+\frac{1}{12}z for x in the equation z=1-x.

$$z=1-\left(\frac{1}{6}+\frac{1}{12}z\right)$$

Solve z=1-\left(\frac{1}{6}+\frac{1}{12}z\right) for z.

$$z=\frac{10}{13}$$

Substitute \frac{10}{13} for z in the equation x=\frac{1}{6}+\frac{1}{12}z.

$$x=\frac{1}{6}+\frac{1}{12}\times \frac{10}{13}$$

Calculate x from x=\frac{1}{6}+\frac{1}{12}\times \frac{10}{13}.

$$x=\frac{3}{13}$$

Substitute \frac{3}{13} for x and \frac{10}{13} for z in the equation y=2x+z+2.

$$y=2\times \frac{3}{13}+\frac{10}{13}+2$$

Calculate y from y=2\times \frac{3}{13}+\frac{10}{13}+2.

$$y=\frac{42}{13}$$

The system is now solved.

$$x=\frac{3}{13} y=\frac{42}{13} z=\frac{10}{13}$$