Solve {l}{x_{1}+x_{2}-x_{3}=0}{2x_{1}-x_{2}-x_{3}=2}{4x_{1}+x_{2}+x_{3}=6}

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

Solve for x_1, x_2, x_3

$$x_{1} = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x_{2}=-\frac{1}{3}\approx -0.333333333$$

$$x_{3}=1$$

Short Steps Using Substitution

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

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

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

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

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

$$x_{2}=-\frac{2}{3}+\frac{1}{3}x_{3} x_{3}=\frac{6}{5}+\frac{3}{5}x_{2}$$

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

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

Solve x_{3}=\frac{6}{5}+\frac{3}{5}\left(-\frac{2}{3}+\frac{1}{3}x_{3}\right) for x_{3}.

$$x_{3}=1$$

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

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

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

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

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

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

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

$$x_{1}=\frac{4}{3}$$

The system is now solved.

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