Solve {c}{x_{1}+quad+6x_{3}=12}{2x_{1}+3x_{2}+9x_{3}=0}{+15x_{2}+9x_{3}=-9}

$$\left. \begin{array} { c } { x _ { 1 } + \quad + 6 x _ { 3 } = 12 } \\ { 2 x _ { 1 } + 3 x _ { 2 } + 9 x _ { 3 } = 0 } \\ { + 15 x _ { 2 } + 9 x _ { 3 } = - 9 } \end{array} \right.$$

Solve for x_1, x_3, x_2

$$x_{1} = -\frac{63}{4} = -15\frac{3}{4} = -15.75$$

$$x_{3} = \frac{37}{8} = 4\frac{5}{8} = 4.625$$

$$x_{2} = -\frac{27}{8} = -3\frac{3}{8} = -3.375$$

Short Steps Using Substitution

Solve x_{1}+6x_{3}=12 for x_{1}.

$$x_{1}=12-6x_{3}$$

Substitute 12-6x_{3} for x_{1} in the equation 2x_{1}+3x_{2}+9x_{3}=0.

$$2\left(12-6x_{3}\right)+3x_{2}+9x_{3}=0$$

Solve the second equation for x_{3} and the third equation for x_{2}.

$$x_{3}=8+x_{2} x_{2}=-\frac{3}{5}-\frac{3}{5}x_{3}$$

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

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

Solve x_{2}=-\frac{3}{5}-\frac{3}{5}\left(8+x_{2}\right) for x_{2}.

$$x_{2}=-\frac{27}{8}$$

Substitute -\frac{27}{8} for x_{2} in the equation x_{3}=8+x_{2}.

$$x_{3}=8-\frac{27}{8}$$

Calculate x_{3} from x_{3}=8-\frac{27}{8}.

$$x_{3}=\frac{37}{8}$$

Substitute \frac{37}{8} for x_{3} in the equation x_{1}=12-6x_{3}.

$$x_{1}=12-6\times \frac{37}{8}$$

Calculate x_{1} from x_{1}=12-6\times \frac{37}{8}.

$$x_{1}=-\frac{63}{4}$$

The system is now solved.

$$x_{1}=-\frac{63}{4} x_{3}=\frac{37}{8} x_{2}=-\frac{27}{8}$$