Solve {r}{2x_{1}-x_{2}+x_{3}=5}{x_{1}+2x_{2}-3x_{3}=6}{3x_{2}-x_{3}=1}

$$\left. \begin{array} { r } { 2 x _ { 1 } - x _ { 2 } + x _ { 3 } = 5 } \\ { x _ { 1 } + 2 x _ { 2 } - 3 x _ { 3 } = 6 } \\ { 3 x _ { 2 } - x _ { 3 } = 1 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=3$$

$$x_{2}=0$$

$$x_{3}=-1$$

Short Steps Using Substitution

Solve 2x_{1}-x_{2}+x_{3}=5 for x_{2}.

$$x_{2}=2x_{1}+x_{3}-5$$

Substitute 2x_{1}+x_{3}-5 for x_{2} in the second and third equation.

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

Solve these equations for x_{1} and x_{3} respectively.

$$x_{1}=\frac{16}{5}+\frac{1}{5}x_{3} x_{3}=8-3x_{1}$$

Substitute \frac{16}{5}+\frac{1}{5}x_{3} for x_{1} in the equation x_{3}=8-3x_{1}.

$$x_{3}=8-3\left(\frac{16}{5}+\frac{1}{5}x_{3}\right)$$

Solve x_{3}=8-3\left(\frac{16}{5}+\frac{1}{5}x_{3}\right) for x_{3}.

$$x_{3}=-1$$

Substitute -1 for x_{3} in the equation x_{1}=\frac{16}{5}+\frac{1}{5}x_{3}.

$$x_{1}=\frac{16}{5}+\frac{1}{5}\left(-1\right)$$

Calculate x_{1} from x_{1}=\frac{16}{5}+\frac{1}{5}\left(-1\right).

$$x_{1}=3$$

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

$$x_{2}=2\times 3-1-5$$

Calculate x_{2} from x_{2}=2\times 3-1-5.

$$x_{2}=0$$

The system is now solved.

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