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

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

Solve for x_1, x_2, x_3

$$x_{1}=1$$

$$x_{2}=-3$$

$$x_{3}=-1$$

Short Steps Using Substitution

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

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

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

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

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

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

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

$$x_{3}=-\left(-\frac{4}{3}x_{3}-\frac{1}{3}\right)$$

Solve x_{3}=-\left(-\frac{4}{3}x_{3}-\frac{1}{3}\right) for x_{3}.

$$x_{3}=-1$$

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

$$x_{1}=-\frac{4}{3}\left(-1\right)-\frac{1}{3}$$

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

$$x_{1}=1$$

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

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

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

$$x_{2}=-3$$

The system is now solved.

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