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

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

Solve for x_1, x_2, x_3

$$x_{1}=2$$

$$x_{2}=-1$$

$$x_{3}=\frac{1}{2}=0.5$$

Short Steps Using Substitution

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

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

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

$$-x_{1}+2\left(2x_{1}-4x_{3}-3\right)+10x_{3}=1 4x_{1}+4\left(2x_{1}-4x_{3}-3\right)-4x_{3}=2$$

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

$$x_{1}=-\frac{2}{3}x_{3}+\frac{7}{3} x_{3}=-\frac{7}{10}+\frac{3}{5}x_{1}$$

Substitute -\frac{2}{3}x_{3}+\frac{7}{3} for x_{1} in the equation x_{3}=-\frac{7}{10}+\frac{3}{5}x_{1}.

$$x_{3}=-\frac{7}{10}+\frac{3}{5}\left(-\frac{2}{3}x_{3}+\frac{7}{3}\right)$$

Solve x_{3}=-\frac{7}{10}+\frac{3}{5}\left(-\frac{2}{3}x_{3}+\frac{7}{3}\right) for x_{3}.

$$x_{3}=\frac{1}{2}$$

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

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

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

$$x_{1}=2$$

Substitute 2 for x_{1} and \frac{1}{2} for x_{3} in the equation x_{2}=2x_{1}-4x_{3}-3.

$$x_{2}=2\times 2-4\times \frac{1}{2}-3$$

Calculate x_{2} from x_{2}=2\times 2-4\times \frac{1}{2}-3.

$$x_{2}=-1$$

The system is now solved.

$$x_{1}=2 x_{2}=-1 x_{3}=\frac{1}{2}$$