Solve {r}{-6x+y+2z=-20}{33x-30y-32z=20}{3x-y-2z=0}

$$\left. \begin{array} { r } { - 6 x + y + 2 z = - 20 } \\ { 33 x - 30 y - 32 z = 20 } \\ { 3 x - y - 2 z = 0 } \end{array} \right.$$

Solve for x, y, z

$$x = \frac{20}{3} = 6\frac{2}{3} \approx 6.666666667$$

$$y = -\frac{60}{7} = -8\frac{4}{7} \approx -8.571428571$$

$$z = \frac{100}{7} = 14\frac{2}{7} \approx 14.285714286$$

Short Steps Using Substitution

Solve -6x+y+2z=-20 for y.

$$y=-20+6x-2z$$

Substitute -20+6x-2z for y in the second and third equation.

$$33x-30\left(-20+6x-2z\right)-32z=20 3x-\left(-20+6x-2z\right)-2z=0$$

Solve these equations for z and x respectively.

$$z=\frac{21}{4}x-\frac{145}{7} x=\frac{20}{3}$$

Substitute \frac{20}{3} for x in the equation z=\frac{21}{4}x-\frac{145}{7}.

$$z=\frac{21}{4}\times \frac{20}{3}-\frac{145}{7}$$

Calculate z from z=\frac{21}{4}\times \frac{20}{3}-\frac{145}{7}.

$$z=\frac{100}{7}$$

Substitute \frac{100}{7} for z and \frac{20}{3} for x in the equation y=-20+6x-2z.

$$y=-20+6\times \frac{20}{3}-2\times \frac{100}{7}$$

Calculate y from y=-20+6\times \frac{20}{3}-2\times \frac{100}{7}.

$$y=-\frac{60}{7}$$

The system is now solved.

$$x=\frac{20}{3} y=-\frac{60}{7} z=\frac{100}{7}$$