Solve {c}{2x+y-3z=-1}{x-3y-2z=-12}{3x-2y-z=-5}

$$\left. \begin{array} { c } { 2 x + y - 3 z = - 1 } \\ { x - 3 y - 2 z = - 12 } \\ { 3 x - 2 y - z = - 5 } \end{array} \right.$$

Solve for x, y, z

$$x=1$$

$$y=3$$

$$z=2$$

Short Steps Using Substitution

Solve 2x+y-3z=-1 for y.

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

Substitute -2x+3z-1 for y in the second and third equation.

$$x-3\left(-2x+3z-1\right)-2z=-12 3x-2\left(-2x+3z-1\right)-z=-5$$

Solve these equations for x and z respectively.

$$x=\frac{11}{7}z-\frac{15}{7} z=1+x$$

Substitute \frac{11}{7}z-\frac{15}{7} for x in the equation z=1+x.

$$z=1+\frac{11}{7}z-\frac{15}{7}$$

Solve z=1+\frac{11}{7}z-\frac{15}{7} for z.

$$z=2$$

Substitute 2 for z in the equation x=\frac{11}{7}z-\frac{15}{7}.

$$x=\frac{11}{7}\times 2-\frac{15}{7}$$

Calculate x from x=\frac{11}{7}\times 2-\frac{15}{7}.

$$x=1$$

Substitute 1 for x and 2 for z in the equation y=-2x+3z-1.

$$y=-2+3\times 2-1$$

Calculate y from y=-2+3\times 2-1.

$$y=3$$

The system is now solved.

$$x=1 y=3 z=2$$