Solve {l}{text{youca.}}{2x_{1}-3x_{2}-x_{3}+6=0}{3x_{1}+4x_{2}+3x_{3}+5=0}{x_{1}+x_{2}+x_{3}+2=0}

$$\left. \begin{array} { l } { \text { youca. } } \\ { 2 x _ { 1 } - 3 x _ { 2 } - x _ { 3 } + 6 = 0 } \\ { 3 x _ { 1 } + 4 x _ { 2 } + 3 x _ { 3 } + 5 = 0 } \\ { x _ { 1 } + x _ { 2 } + x _ { 3 } + 2 = 0 } \end{array} \right.$$

Solve for x_2, x_3, x_1

$$x_{2}=1$$

$$x_{3}=3$$

$$x_{1}=-6$$

Short Steps Using Substitution

Solve -3x_{2}-x_{3}+6=0 for x_{3}.

$$x_{3}=-3x_{2}+6$$

Substitute -3x_{2}+6 for x_{3} in the second and third equation.

$$3x_{1}+4x_{2}+3\left(-3x_{2}+6\right)+5=0 x_{1}+x_{2}-3x_{2}+6+2=0$$

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

$$x_{2}=\frac{3}{5}x_{1}+\frac{23}{5} x_{1}=-8+2x_{2}$$

Substitute \frac{3}{5}x_{1}+\frac{23}{5} for x_{2} in the equation x_{1}=-8+2x_{2}.

$$x_{1}=-8+2\left(\frac{3}{5}x_{1}+\frac{23}{5}\right)$$

Solve x_{1}=-8+2\left(\frac{3}{5}x_{1}+\frac{23}{5}\right) for x_{1}.

$$x_{1}=-6$$

Substitute -6 for x_{1} in the equation x_{2}=\frac{3}{5}x_{1}+\frac{23}{5}.

$$x_{2}=\frac{3}{5}\left(-6\right)+\frac{23}{5}$$

Calculate x_{2} from x_{2}=\frac{3}{5}\left(-6\right)+\frac{23}{5}.

$$x_{2}=1$$

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

$$x_{3}=-3+6$$

Calculate x_{3} from x_{3}=-3+6.

$$x_{3}=3$$

The system is now solved.

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