Solve {r}{-4x_{1}+1x_{2}-2x_{3}=2}{x_{1}-2x_{2}+x_{3}=3}{-3x_{1}+5x_{2}-x_{3}=-4}

$$\left. \begin{array} { r } { - 4 x _ { 1 } + 1 x _ { 2 } - 2 x _ { 3 } = 2 } \\ { x _ { 1 } - 2 x _ { 2 } + x _ { 3 } = 3 } \\ { - 3 x _ { 1 } + 5 x _ { 2 } - x _ { 3 } = - 4 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = -\frac{7}{4} = -1\frac{3}{4} = -1.75$$

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

$$x_{3} = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

Short Steps Using Substitution

Reorder the equations.

$$x_{1}-2x_{2}+x_{3}=3 -4x_{1}+1x_{2}-2x_{3}=2 -3x_{1}+5x_{2}-x_{3}=-4$$

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

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

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

$$-4\left(2x_{2}-x_{3}+3\right)+1x_{2}-2x_{3}=2 -3\left(2x_{2}-x_{3}+3\right)+5x_{2}-x_{3}=-4$$

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

$$x_{2}=-2+\frac{2}{7}x_{3} x_{3}=\frac{5}{2}+\frac{1}{2}x_{2}$$

Substitute -2+\frac{2}{7}x_{3} for x_{2} in the equation x_{3}=\frac{5}{2}+\frac{1}{2}x_{2}.

$$x_{3}=\frac{5}{2}+\frac{1}{2}\left(-2+\frac{2}{7}x_{3}\right)$$

Solve x_{3}=\frac{5}{2}+\frac{1}{2}\left(-2+\frac{2}{7}x_{3}\right) for x_{3}.

$$x_{3}=\frac{7}{4}$$

Substitute \frac{7}{4} for x_{3} in the equation x_{2}=-2+\frac{2}{7}x_{3}.

$$x_{2}=-2+\frac{2}{7}\times \frac{7}{4}$$

Calculate x_{2} from x_{2}=-2+\frac{2}{7}\times \frac{7}{4}.

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

Substitute -\frac{3}{2} for x_{2} and \frac{7}{4} for x_{3} in the equation x_{1}=2x_{2}-x_{3}+3.

$$x_{1}=2\left(-\frac{3}{2}\right)-\frac{7}{4}+3$$

Calculate x_{1} from x_{1}=2\left(-\frac{3}{2}\right)-\frac{7}{4}+3.

$$x_{1}=-\frac{7}{4}$$

The system is now solved.

$$x_{1}=-\frac{7}{4} x_{2}=-\frac{3}{2} x_{3}=\frac{7}{4}$$