Solve {l}{3x-y+z=30}{2x+y-3z=11θ}{x+y+z=2}

$$\left\{ \begin{array} { l } { 3 x - y + z = 30 } \\ { 2 x + y - 3 z = 11 \theta } \\ { x + y + z = 2 } \end{array} \right.$$

Solve for x, y, z

$$x=\frac{11\theta +62}{9}$$

$$y=\frac{11\theta -64}{9}$$

$$z=\frac{20-22\theta }{9}$$

Short Steps Using Substitution

Solve 3x-y+z=30 for y.

$$y=3x+z-30$$

Substitute 3x+z-30 for y in the second and third equation.

$$2x+3x+z-30-3z=11\theta x+3x+z-30+z=2$$

Solve these equations for x and z respectively.

$$x=\frac{2}{5}z+6+\frac{11}{5}\theta z=16-2x$$

Substitute \frac{2}{5}z+6+\frac{11}{5}\theta for x in the equation z=16-2x.

$$z=16-2\left(\frac{2}{5}z+6+\frac{11}{5}\theta \right)$$

Solve z=16-2\left(\frac{2}{5}z+6+\frac{11}{5}\theta \right) for z.

$$z=\frac{20}{9}-\frac{22}{9}\theta$$

Substitute \frac{20}{9}-\frac{22}{9}\theta for z in the equation x=\frac{2}{5}z+6+\frac{11}{5}\theta .

$$x=\frac{2}{5}\left(\frac{20}{9}-\frac{22}{9}\theta \right)+6+\frac{11}{5}\theta$$

Calculate x from x=\frac{2}{5}\left(\frac{20}{9}-\frac{22}{9}\theta \right)+6+\frac{11}{5}\theta .

$$x=\frac{62}{9}+\frac{11}{9}\theta$$

Substitute \frac{62}{9}+\frac{11}{9}\theta for x and \frac{20}{9}-\frac{22}{9}\theta for z in the equation y=3x+z-30.

$$y=3\left(\frac{62}{9}+\frac{11}{9}\theta \right)+\frac{20}{9}-\frac{22}{9}\theta -30$$

Calculate y from y=3\left(\frac{62}{9}+\frac{11}{9}\theta \right)+\frac{20}{9}-\frac{22}{9}\theta -30.

$$y=-\frac{64}{9}+\frac{11}{9}\theta$$

The system is now solved.

$$x=\frac{62}{9}+\frac{11}{9}\theta y=-\frac{64}{9}+\frac{11}{9}\theta z=\frac{20}{9}-\frac{22}{9}\theta$$