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

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

Solve for x_1, x_2, x_3

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

$$x_{2}=\frac{-x_{4}-5}{2}$$

$$x_{3}=-2x_{4}-9$$

Short Steps Using Substitution

Reorder the equations.

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

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

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

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

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

Solve the second equation for x_{3} and the third equation for x_{2}.

$$x_{3}=4x_{2}+1 x_{2}=-\frac{5}{2}-\frac{1}{2}x_{4}$$

Substitute -\frac{5}{2}-\frac{1}{2}x_{4} for x_{2} in the equation x_{3}=4x_{2}+1.

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

Calculate x_{3} from x_{3}=4\left(-\frac{5}{2}-\frac{1}{2}x_{4}\right)+1.

$$x_{3}=-9-2x_{4}$$

Substitute -\frac{5}{2}-\frac{1}{2}x_{4} for x_{2} in the equation x_{1}=-x_{2}-3.

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

Calculate x_{1} from x_{1}=-\left(-\frac{5}{2}-\frac{1}{2}x_{4}\right)-3.

$$x_{1}=-\frac{1}{2}+\frac{1}{2}x_{4}$$

The system is now solved.

$$x_{1}=-\frac{1}{2}+\frac{1}{2}x_{4} x_{2}=-\frac{5}{2}-\frac{1}{2}x_{4} x_{3}=-9-2x_{4}$$