Solve {l}{-2x_{1}+3x_{2}-x_{3}=1}{x_{1}+2x_{2}-x_{3}=-4}{-2x_{1}-x_{2}+x_{3}=-3}

$$\left. \begin{array}{l}{ - 2 x _ { 1 } + 3 x _ { 2 } - x _ { 3 } = 1 }\\{ x _ { 1 } + 2 x _ { 2 } - x _ { 3 } = - 4 }\\{ - 2 x _ { 1 } - x _ { 2 } + x _ { 3 } = - 3 }\end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=-6$$

$$x_{2}=-13$$

$$x_{3}=-28$$

Short Steps Using Substitution

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

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

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

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

Solve these equations for x_{2} and x_{1} respectively.

$$x_{2}=3x_{1}+5 x_{1}=\frac{1}{2}+\frac{1}{2}x_{2}$$

Substitute 3x_{1}+5 for x_{2} in the equation x_{1}=\frac{1}{2}+\frac{1}{2}x_{2}.

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

Solve x_{1}=\frac{1}{2}+\frac{1}{2}\left(3x_{1}+5\right) for x_{1}.

$$x_{1}=-6$$

Substitute -6 for x_{1} in the equation x_{2}=3x_{1}+5.

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

Calculate x_{2} from x_{2}=3\left(-6\right)+5.

$$x_{2}=-13$$

Substitute -13 for x_{2} and -6 for x_{1} in the equation x_{3}=-2x_{1}+3x_{2}-1.

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

Calculate x_{3} from x_{3}=-2\left(-6\right)+3\left(-13\right)-1.

$$x_{3}=-28$$

The system is now solved.

$$x_{1}=-6 x_{2}=-13 x_{3}=-28$$