Solve {r}{2x_{1}+3x_{2}+x_{3}=4}{x_{1}+2x_{2}-2x_{3}=7}{2x_{1}-5x_{2}+3x_{3}=0}

$$\left. \begin{array} { r } { 2 x _ { 1 } + 3 x _ { 2 } + x _ { 3 } = 4 } \\ { x _ { 1 } + 2 x _ { 2 } - 2 x _ { 3 } = 7 } \\ { 2 x _ { 1 } - 5 x _ { 2 } + 3 x _ { 3 } = 0 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1}=3$$

$$x_{2}=0$$

$$x_{3}=-2$$

Short Steps Using Substitution

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

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

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

$$x_{1}+2x_{2}-2\left(-2x_{1}-3x_{2}+4\right)=7 2x_{1}-5x_{2}+3\left(-2x_{1}-3x_{2}+4\right)=0$$

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

$$x_{2}=\frac{15}{8}-\frac{5}{8}x_{1} x_{1}=3-\frac{7}{2}x_{2}$$

Substitute \frac{15}{8}-\frac{5}{8}x_{1} for x_{2} in the equation x_{1}=3-\frac{7}{2}x_{2}.

$$x_{1}=3-\frac{7}{2}\left(\frac{15}{8}-\frac{5}{8}x_{1}\right)$$

Solve x_{1}=3-\frac{7}{2}\left(\frac{15}{8}-\frac{5}{8}x_{1}\right) for x_{1}.

$$x_{1}=3$$

Substitute 3 for x_{1} in the equation x_{2}=\frac{15}{8}-\frac{5}{8}x_{1}.

$$x_{2}=\frac{15}{8}-\frac{5}{8}\times 3$$

Calculate x_{2} from x_{2}=\frac{15}{8}-\frac{5}{8}\times 3.

$$x_{2}=0$$

Substitute 0 for x_{2} and 3 for x_{1} in the equation x_{3}=-2x_{1}-3x_{2}+4.

$$x_{3}=-2\times 3-3\times 0+4$$

Calculate x_{3} from x_{3}=-2\times 3-3\times 0+4.

$$x_{3}=-2$$

The system is now solved.

$$x_{1}=3 x_{2}=0 x_{3}=-2$$