Solve {l}{5x_{1}-8x_{2}-4x_{3}-1=0}{2x_{1}+x_{2}+5x_{3}-4=0}{3x_{3}-2x_{2}+2x_{3}-2=0}

$$\left. \begin{array} { l } { 5 x _ { 1 } - 8 x _ { 2 } - 4 x _ { 3 } - 1 = 0 } \\ { 2 x _ { 1 } + x _ { 2 } + 5 x _ { 3 } - 4 = 0 } \\ { 3 x _ { 3 } - 2 x _ { 2 } + 2 x _ { 3 } - 2 = 0 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=\frac{15}{19}\approx 0.789473684$$

$$x_{2}=\frac{8}{57}\approx 0.140350877$$

$$x_{3}=\frac{26}{57}\approx 0.456140351$$

Short Steps Using Substitution

Reorder the equations.

$$2x_{1}+x_{2}+5x_{3}-4=0 5x_{1}-8x_{2}-4x_{3}-1=0 3x_{3}-2x_{2}+2x_{3}-2=0$$

Solve 2x_{1}+x_{2}+5x_{3}-4=0 for x_{2}.

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

Substitute -2x_{1}-5x_{3}+4 for x_{2} in the second and third equation.

$$5x_{1}-8\left(-2x_{1}-5x_{3}+4\right)-4x_{3}-1=0 3x_{3}-2\left(-2x_{1}-5x_{3}+4\right)+2x_{3}-2=0$$

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

$$x_{1}=\frac{11}{7}-\frac{12}{7}x_{3} x_{3}=\frac{2}{3}-\frac{4}{15}x_{1}$$

Substitute \frac{11}{7}-\frac{12}{7}x_{3} for x_{1} in the equation x_{3}=\frac{2}{3}-\frac{4}{15}x_{1}.

$$x_{3}=\frac{2}{3}-\frac{4}{15}\left(\frac{11}{7}-\frac{12}{7}x_{3}\right)$$

Solve x_{3}=\frac{2}{3}-\frac{4}{15}\left(\frac{11}{7}-\frac{12}{7}x_{3}\right) for x_{3}.

$$x_{3}=\frac{26}{57}$$

Substitute \frac{26}{57} for x_{3} in the equation x_{1}=\frac{11}{7}-\frac{12}{7}x_{3}.

$$x_{1}=\frac{11}{7}-\frac{12}{7}\times \frac{26}{57}$$

Calculate x_{1} from x_{1}=\frac{11}{7}-\frac{12}{7}\times \frac{26}{57}.

$$x_{1}=\frac{15}{19}$$

Substitute \frac{15}{19} for x_{1} and \frac{26}{57} for x_{3} in the equation x_{2}=-2x_{1}-5x_{3}+4.

$$x_{2}=-2\times \frac{15}{19}-5\times \frac{26}{57}+4$$

Calculate x_{2} from x_{2}=-2\times \frac{15}{19}-5\times \frac{26}{57}+4.

$$x_{2}=\frac{8}{57}$$

The system is now solved.

$$x_{1}=\frac{15}{19} x_{2}=\frac{8}{57} x_{3}=\frac{26}{57}$$