Solve {r}{9x_{1}+9x_{2}-7x_{3}=6}{-7x_{1}+6x_{2}-x_{3}=-10}{9x_{1}+6x_{2}+8x_{3}=45}

$$\left. \begin{array} { r } { 9 x _ { 1 } + 9 x _ { 2 } - 7 x _ { 3 } = 6 } \\ { - 7 x _ { 1 } + 6 x _ { 2 } - x _ { 3 } = - 10 } \\ { 9 x _ { 1 } + 6 x _ { 2 } + 8 x _ { 3 } = 45 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{983}{527} = 1\frac{456}{527} \approx 1.865275142$$

$$x_{2}=\frac{514}{527}\approx 0.975332068$$

$$x_{3} = \frac{1473}{527} = 2\frac{419}{527} \approx 2.795066414$$

Short Steps Using Substitution

Reorder the equations.

$$-7x_{1}+6x_{2}-x_{3}=-10 9x_{1}+9x_{2}-7x_{3}=6 9x_{1}+6x_{2}+8x_{3}=45$$

Solve -7x_{1}+6x_{2}-x_{3}=-10 for x_{3}.

$$x_{3}=-7x_{1}+6x_{2}+10$$

Substitute -7x_{1}+6x_{2}+10 for x_{3} in the second and third equation.

$$9x_{1}+9x_{2}-7\left(-7x_{1}+6x_{2}+10\right)=6 9x_{1}+6x_{2}+8\left(-7x_{1}+6x_{2}+10\right)=45$$

Solve these equations for x_{2} and x_{1} respectively.

$$x_{2}=-\frac{76}{33}+\frac{58}{33}x_{1} x_{1}=\frac{54}{47}x_{2}+\frac{35}{47}$$

Substitute -\frac{76}{33}+\frac{58}{33}x_{1} for x_{2} in the equation x_{1}=\frac{54}{47}x_{2}+\frac{35}{47}.

$$x_{1}=\frac{54}{47}\left(-\frac{76}{33}+\frac{58}{33}x_{1}\right)+\frac{35}{47}$$

Solve x_{1}=\frac{54}{47}\left(-\frac{76}{33}+\frac{58}{33}x_{1}\right)+\frac{35}{47} for x_{1}.

$$x_{1}=\frac{983}{527}$$

Substitute \frac{983}{527} for x_{1} in the equation x_{2}=-\frac{76}{33}+\frac{58}{33}x_{1}.

$$x_{2}=-\frac{76}{33}+\frac{58}{33}\times \frac{983}{527}$$

Calculate x_{2} from x_{2}=-\frac{76}{33}+\frac{58}{33}\times \frac{983}{527}.

$$x_{2}=\frac{514}{527}$$

Substitute \frac{514}{527} for x_{2} and \frac{983}{527} for x_{1} in the equation x_{3}=-7x_{1}+6x_{2}+10.

$$x_{3}=-7\times \frac{983}{527}+6\times \frac{514}{527}+10$$

Calculate x_{3} from x_{3}=-7\times \frac{983}{527}+6\times \frac{514}{527}+10.

$$x_{3}=\frac{1473}{527}$$

The system is now solved.

$$x_{1}=\frac{983}{527} x_{2}=\frac{514}{527} x_{3}=\frac{1473}{527}$$