Solve {l}{x_{1}+x_{2}+2x_{3}=8}{x_{1}+2x_{2}-3x_{3}=-1}{3x_{1}-7x_{2}=4x_{3}=10}

$$\left. \begin{array} { l } { x _ { 1 } + x _ { 2 } + 2 x _ { 3 } = 8 } \\ { x _ { 1 } + 2 x _ { 2 } - 3 x _ { 3 } = - 1 } \\ { 3 x _ { 1 } - 7 x _ { 2 } = 4 x _ { 3 } = 10 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{37}{10} = 3\frac{7}{10} = 3.7$$

$$x_{2}=\frac{1}{2}=0.5$$

$$x_{3} = \frac{19}{10} = 1\frac{9}{10} = 1.9$$

Short Steps Using Substitution

Solve x_{1}+x_{2}+2x_{3}=8 for x_{1}.

$$x_{1}=-x_{2}-2x_{3}+8$$

Substitute -x_{2}-2x_{3}+8 for x_{1} in the second and third equation.

$$-x_{2}-2x_{3}+8+2x_{2}-3x_{3}=-1 3\left(-x_{2}-2x_{3}+8\right)-7x_{2}=4x_{3}$$

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

$$x_{2}=5x_{3}-9 x_{3}=\frac{12}{5}-x_{2}$$

Substitute 5x_{3}-9 for x_{2} in the equation x_{3}=\frac{12}{5}-x_{2}.

$$x_{3}=\frac{12}{5}-\left(5x_{3}-9\right)$$

Solve x_{3}=\frac{12}{5}-\left(5x_{3}-9\right) for x_{3}.

$$x_{3}=\frac{19}{10}$$

Substitute \frac{19}{10} for x_{3} in the equation x_{2}=5x_{3}-9.

$$x_{2}=5\times \frac{19}{10}-9$$

Calculate x_{2} from x_{2}=5\times \frac{19}{10}-9.

$$x_{2}=\frac{1}{2}$$

Substitute \frac{1}{2} for x_{2} and \frac{19}{10} for x_{3} in the equation x_{1}=-x_{2}-2x_{3}+8.

$$x_{1}=-\frac{1}{2}-2\times \frac{19}{10}+8$$

Calculate x_{1} from x_{1}=-\frac{1}{2}-2\times \frac{19}{10}+8.

$$x_{1}=\frac{37}{10}$$

The system is now solved.

$$x_{1}=\frac{37}{10} x_{2}=\frac{1}{2} x_{3}=\frac{19}{10}$$