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

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

Solve for x_1, x_2, x_3

$$x_{1}=2$$

$$x_{2}=3$$

$$x_{3}=4$$

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.

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

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

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

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

$$x_{1}=2$$

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

$$x_{2}=-3+3\times 2$$

Calculate x_{2} from x_{2}=-3+3\times 2.

$$x_{2}=3$$

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

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

Calculate x_{3} from x_{3}=-2\times 2+3\times 3-1.

$$x_{3}=4$$

The system is now solved.

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