Solve {c}{x_{1}+3x_{2}+2x_{3}=2}{3x_{1}+8x_{2}+6x_{3}=-1}{2x_{1}+7x_{2}+3x_{3}=1}

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

Solve for x_1, x_2, x_3

$$x_{1}=-39$$

$$x_{2}=7$$

$$x_{3}=10$$

Short Steps Using Substitution

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

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

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

$$3\left(-3x_{2}-2x_{3}+2\right)+8x_{2}+6x_{3}=-1 2\left(-3x_{2}-2x_{3}+2\right)+7x_{2}+3x_{3}=1$$

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

$$x_{2}=7 x_{3}=x_{2}+3$$

Substitute 7 for x_{2} in the equation x_{3}=x_{2}+3.

$$x_{3}=7+3$$

Calculate x_{3} from x_{3}=7+3.

$$x_{3}=10$$

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

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

Calculate x_{1} from x_{1}=-3\times 7-2\times 10+2.

$$x_{1}=-39$$

The system is now solved.

$$x_{1}=-39 x_{2}=7 x_{3}=10$$