Solve {r}{x_{1}+x_{2}-3x_{3}-2x_{4}=2}{x_{1}-x_{3}-x_{4}=1}{2x_{1}+6x_{2}-13x_{3}-5x_{4}=6}

$$\left. \begin{array} { r } { x _ { 1 } + x _ { 2 } - 3 x _ { 3 } - 2 x _ { 4 } = 2 } \\ { x _ { 1 } - x _ { 3 } - x _ { 4 } = 1 } \\ { 2 x _ { 1 } + 6 x _ { 2 } - 13 x _ { 3 } - 5 x _ { 4 } = 6 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=-2x_{4}-1$$

$$x_{2}=-5x_{4}-3$$

$$x_{3}=-3x_{4}-2$$

Short Steps Using Substitution

Solve x_{1}+x_{2}-3x_{3}-2x_{4}=2 for x_{1}.

$$x_{1}=-x_{2}+3x_{3}+2x_{4}+2$$

Substitute -x_{2}+3x_{3}+2x_{4}+2 for x_{1} in the second and third equation.

$$-x_{2}+3x_{3}+2x_{4}+2-x_{3}-x_{4}=1 2\left(-x_{2}+3x_{3}+2x_{4}+2\right)+6x_{2}-13x_{3}-5x_{4}=6$$

Solve these equations for x_{2} and x_{3} respectively.

$$x_{2}=2x_{3}+x_{4}+1 x_{3}=\frac{4}{7}x_{2}-\frac{1}{7}x_{4}-\frac{2}{7}$$

Substitute 2x_{3}+x_{4}+1 for x_{2} in the equation x_{3}=\frac{4}{7}x_{2}-\frac{1}{7}x_{4}-\frac{2}{7}.

$$x_{3}=\frac{4}{7}\left(2x_{3}+x_{4}+1\right)-\frac{1}{7}x_{4}-\frac{2}{7}$$

Solve x_{3}=\frac{4}{7}\left(2x_{3}+x_{4}+1\right)-\frac{1}{7}x_{4}-\frac{2}{7} for x_{3}.

$$x_{3}=-2-3x_{4}$$

Substitute -2-3x_{4} for x_{3} in the equation x_{2}=2x_{3}+x_{4}+1.

$$x_{2}=2\left(-2-3x_{4}\right)+x_{4}+1$$

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

$$x_{2}=-3-5x_{4}$$

Substitute -3-5x_{4} for x_{2} and -2-3x_{4} for x_{3} in the equation x_{1}=-x_{2}+3x_{3}+2x_{4}+2.

$$x_{1}=-\left(-3-5x_{4}\right)+3\left(-2-3x_{4}\right)+2x_{4}+2$$

Calculate x_{1} from x_{1}=-\left(-3-5x_{4}\right)+3\left(-2-3x_{4}\right)+2x_{4}+2.

$$x_{1}=-1-2x_{4}$$

The system is now solved.

$$x_{1}=-1-2x_{4} x_{2}=-3-5x_{4} x_{3}=-2-3x_{4}$$