Solve {l}{3x_{1}+2x_{2}+4x_{3}=-3}{2x_{1}+4x_{2}+x_{3}=10}{x_{1}+x_{2}-x_{3}=9}

$$\left\{ \begin{array} { l } { 3 x _ { 1 } + 2 x _ { 2 } + 4 x _ { 3 } = - 3 } \\ { 2 x _ { 1 } + 4 x _ { 2 } + x _ { 3 } = 10 } \\ { x _ { 1 } + x _ { 2 } - x _ { 3 } = 9 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=3$$

$$x_{2}=2$$

$$x_{3}=-4$$

Short Steps Using Substitution

Reorder the equations.

$$2x_{1}+4x_{2}+x_{3}=10 3x_{1}+2x_{2}+4x_{3}=-3 x_{1}+x_{2}-x_{3}=9$$

Solve 2x_{1}+4x_{2}+x_{3}=10 for x_{3}.

$$x_{3}=10-2x_{1}-4x_{2}$$

Substitute 10-2x_{1}-4x_{2} for x_{3} in the second and third equation.

$$3x_{1}+2x_{2}+4\left(10-2x_{1}-4x_{2}\right)=-3 x_{1}+x_{2}-\left(10-2x_{1}-4x_{2}\right)=9$$

Solve these equations for x_{2} and x_{1} respectively.

$$x_{2}=-\frac{5}{14}x_{1}+\frac{43}{14} x_{1}=-\frac{5}{3}x_{2}+\frac{19}{3}$$

Substitute -\frac{5}{14}x_{1}+\frac{43}{14} for x_{2} in the equation x_{1}=-\frac{5}{3}x_{2}+\frac{19}{3}.

$$x_{1}=-\frac{5}{3}\left(-\frac{5}{14}x_{1}+\frac{43}{14}\right)+\frac{19}{3}$$

Solve x_{1}=-\frac{5}{3}\left(-\frac{5}{14}x_{1}+\frac{43}{14}\right)+\frac{19}{3} for x_{1}.

$$x_{1}=3$$

Substitute 3 for x_{1} in the equation x_{2}=-\frac{5}{14}x_{1}+\frac{43}{14}.

$$x_{2}=-\frac{5}{14}\times 3+\frac{43}{14}$$

Calculate x_{2} from x_{2}=-\frac{5}{14}\times 3+\frac{43}{14}.

$$x_{2}=2$$

Substitute 2 for x_{2} and 3 for x_{1} in the equation x_{3}=10-2x_{1}-4x_{2}.

$$x_{3}=10-2\times 3-4\times 2$$

Calculate x_{3} from x_{3}=10-2\times 3-4\times 2.

$$x_{3}=-4$$

The system is now solved.

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