Solve {c}{3x_{1}-x_{2}=1}{-x_{1}+x_{2}-x_{3}=3}{-x_{2}+x_{3}=4}

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

Solve for x_1, x_2, x_3

$$x_{1}=-7$$

$$x_{2}=-22$$

$$x_{3}=-18$$

Short Steps Using Substitution

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

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

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

$$-x_{1}+3x_{1}-1-x_{3}=3 -\left(3x_{1}-1\right)+x_{3}=4$$

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

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

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

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

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

$$x_{3}=-18$$

Substitute -18 for x_{3} in the equation x_{1}=2+\frac{1}{2}x_{3}.

$$x_{1}=2+\frac{1}{2}\left(-18\right)$$

Calculate x_{1} from x_{1}=2+\frac{1}{2}\left(-18\right).

$$x_{1}=-7$$

Substitute -7 for x_{1} in the equation x_{2}=3x_{1}-1.

$$x_{2}=3\left(-7\right)-1$$

Calculate x_{2} from x_{2}=3\left(-7\right)-1.

$$x_{2}=-22$$

The system is now solved.

$$x_{1}=-7 x_{2}=-22 x_{3}=-18$$