Solve {l}{x_{1}-2x_{2}+3x_{3}=0}{4x_{1}+x_{2}-x_{3}=0}{2x_{1}-x_{2}+3x_{3}=0}

$$\left. \begin{array} { l } { x _ { 1 } - 2 x _ { 2 } + 3 x _ { 3 } = 0 } \\ { 4 x _ { 1 } + x _ { 2 } - x _ { 3 } = 0 } \\ { 2 x _ { 1 } - x _ { 2 } + 3 x _ { 3 } = 0 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=0$$

$$x_{2}=0$$

$$x_{3}=0$$

Short Steps Using Substitution

Solve x_{1}-2x_{2}+3x_{3}=0 for x_{1}.

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

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

$$4\left(2x_{2}-3x_{3}\right)+x_{2}-x_{3}=0 2\left(2x_{2}-3x_{3}\right)-x_{2}+3x_{3}=0$$

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

$$x_{2}=\frac{13}{9}x_{3} x_{3}=x_{2}$$

Substitute \frac{13}{9}x_{3} for x_{2} in the equation x_{3}=x_{2}.

$$x_{3}=\frac{13}{9}x_{3}$$

Solve x_{3}=\frac{13}{9}x_{3} for x_{3}.

$$x_{3}=0$$

Substitute 0 for x_{3} in the equation x_{2}=\frac{13}{9}x_{3}.

$$x_{2}=\frac{13}{9}\times 0$$

Calculate x_{2} from x_{2}=\frac{13}{9}\times 0.

$$x_{2}=0$$

Substitute 0 for x_{2} and 0 for x_{3} in the equation x_{1}=2x_{2}-3x_{3}.

$$x_{1}=2\times 0-3\times 0$$

Calculate x_{1} from x_{1}=2\times 0-3\times 0.

$$x_{1}=0$$

The system is now solved.

$$x_{1}=0 x_{2}=0 x_{3}=0$$