Solve {l}{x_{1}+2x_{2}+x_{3}=4}{3x_{1}-4x_{2}-2x_{3}=2}{5x_{1}+3x_{2}+5x_{3}=-1}

$$\left. \begin{array}{l}{ x _ { 1 } + 2 x _ { 2 } + x _ { 3 } = 4 }\\{ 3 x _ { 1 } - 4 x _ { 2 } - 2 x _ { 3 } = 2 }\\{ 5 x _ { 1 } + 3 x _ { 2 } + 5 x _ { 3 } = - 1 }\end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=2$$

$$x_{2}=3$$

$$x_{3}=-4$$

Short Steps Using Substitution

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

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

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

$$3\left(-2x_{2}-x_{3}+4\right)-4x_{2}-2x_{3}=2 5\left(-2x_{2}-x_{3}+4\right)+3x_{2}+5x_{3}=-1$$

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

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

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

$$x_{3}=2-2\times 3$$

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

$$x_{3}=-4$$

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

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

Calculate x_{1} from x_{1}=-2\times 3-\left(-4\right)+4.

$$x_{1}=2$$

The system is now solved.

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