Solve {l}{x_{2}-4x_{3}=8}{2x_{1}-3x_{2}+2x_{3}=1}{4x_{1}-3x_{2}+12x_{3}=1}

$$\left. \begin{array} { l } { x _ { 2 } - 4 x _ { 3 } = 8 } \\ { 2 x _ { 1 } - 3 x _ { 2 } + 2 x _ { 3 } = 1 } \\ { 4 x _ { 1 } - 3 x _ { 2 } + 12 x _ { 3 } = 1 } \end{array} \right.$$

Solve for x_2, x_3, x_1

$$x_{2}=3$$

$$x_{3} = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$x_{1} = \frac{25}{4} = 6\frac{1}{4} = 6.25$$

Short Steps Using Substitution

Solve x_{2}-4x_{3}=8 for x_{2}.

$$x_{2}=8+4x_{3}$$

Substitute 8+4x_{3} for x_{2} in the second and third equation.

$$2x_{1}-3\left(8+4x_{3}\right)+2x_{3}=1 4x_{1}-3\left(8+4x_{3}\right)+12x_{3}=1$$

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

$$x_{3}=\frac{1}{5}x_{1}-\frac{5}{2} x_{1}=\frac{25}{4}$$

Substitute \frac{25}{4} for x_{1} in the equation x_{3}=\frac{1}{5}x_{1}-\frac{5}{2}.

$$x_{3}=\frac{1}{5}\times \frac{25}{4}-\frac{5}{2}$$

Calculate x_{3} from x_{3}=\frac{1}{5}\times \frac{25}{4}-\frac{5}{2}.

$$x_{3}=-\frac{5}{4}$$

Substitute -\frac{5}{4} for x_{3} in the equation x_{2}=8+4x_{3}.

$$x_{2}=8+4\left(-\frac{5}{4}\right)$$

Calculate x_{2} from x_{2}=8+4\left(-\frac{5}{4}\right).

$$x_{2}=3$$

The system is now solved.

$$x_{2}=3 x_{3}=-\frac{5}{4} x_{1}=\frac{25}{4}$$