Solve {c}{x_{1}-2x_{2}+x_{3}=1}{x_{1}+x_{2}+x_{3}=5}{-x_{1}+x_{2}+x_{3}=2}

$$\left\{ \begin{array} { c } { x _ { 1 } - 2 x _ { 2 } + x _ { 3 } = 1 } \\ { x _ { 1 } + x _ { 2 } + x _ { 3 } = 5 } \\ { - x _ { 1 } + x _ { 2 } + x _ { 3 } = 2 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$x_{2} = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x_{3} = \frac{13}{6} = 2\frac{1}{6} \approx 2.166666667$$

Short Steps Using Substitution

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

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

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

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

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

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

Substitute \frac{4}{3} for x_{2} in the equation x_{3}=\frac{3}{2}+\frac{1}{2}x_{2}.

$$x_{3}=\frac{3}{2}+\frac{1}{2}\times \frac{4}{3}$$

Calculate x_{3} from x_{3}=\frac{3}{2}+\frac{1}{2}\times \frac{4}{3}.

$$x_{3}=\frac{13}{6}$$

Substitute \frac{4}{3} for x_{2} and \frac{13}{6} for x_{3} in the equation x_{1}=2x_{2}-x_{3}+1.

$$x_{1}=2\times \frac{4}{3}-\frac{13}{6}+1$$

Calculate x_{1} from x_{1}=2\times \frac{4}{3}-\frac{13}{6}+1.

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

The system is now solved.

$$x_{1}=\frac{3}{2} x_{2}=\frac{4}{3} x_{3}=\frac{13}{6}$$