Solve {l}{2x_{1}+2x_{2}-3x_{3}=4}{5x_{1}-7x_{2}+10x_{3}=11}{3x_{1}-20x_{2}+3x_{3}=-7}

$$\left. \begin{array} { l } { 2 x _ { 1 } + 2 x _ { 2 } - 3 x _ { 3 } = 4 } \\ { 5 x _ { 1 } - 7 x _ { 2 } + 10 x _ { 3 } = 11 } \\ { 3 x _ { 1 } - 20 x _ { 2 } + 3 x _ { 3 } = - 7 } \end{array} \right.$$

Solve for x_1, x_2, x_3

$$x_{1} = \frac{1317}{625} = 2\frac{67}{625} = 2.1072$$

$$x_{2}=\frac{94}{125}=0.752$$

$$x_{3}=\frac{358}{625}=0.5728$$

Short Steps Using Substitution

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

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

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

$$5\left(-x_{2}+\frac{3}{2}x_{3}+2\right)-7x_{2}+10x_{3}=11 3\left(-x_{2}+\frac{3}{2}x_{3}+2\right)-20x_{2}+3x_{3}=-7$$

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

$$x_{2}=-\frac{1}{12}+\frac{35}{24}x_{3} x_{3}=-\frac{26}{15}+\frac{46}{15}x_{2}$$

Substitute -\frac{1}{12}+\frac{35}{24}x_{3} for x_{2} in the equation x_{3}=-\frac{26}{15}+\frac{46}{15}x_{2}.

$$x_{3}=-\frac{26}{15}+\frac{46}{15}\left(-\frac{1}{12}+\frac{35}{24}x_{3}\right)$$

Solve x_{3}=-\frac{26}{15}+\frac{46}{15}\left(-\frac{1}{12}+\frac{35}{24}x_{3}\right) for x_{3}.

$$x_{3}=\frac{358}{625}$$

Substitute \frac{358}{625} for x_{3} in the equation x_{2}=-\frac{1}{12}+\frac{35}{24}x_{3}.

$$x_{2}=-\frac{1}{12}+\frac{35}{24}\times \frac{358}{625}$$

Calculate x_{2} from x_{2}=-\frac{1}{12}+\frac{35}{24}\times \frac{358}{625}.

$$x_{2}=\frac{94}{125}$$

Substitute \frac{94}{125} for x_{2} and \frac{358}{625} for x_{3} in the equation x_{1}=-x_{2}+\frac{3}{2}x_{3}+2.

$$x_{1}=-\frac{94}{125}+\frac{3}{2}\times \frac{358}{625}+2$$

Calculate x_{1} from x_{1}=-\frac{94}{125}+\frac{3}{2}\times \frac{358}{625}+2.

$$x_{1}=\frac{1317}{625}$$

The system is now solved.

$$x_{1}=\frac{1317}{625} x_{2}=\frac{94}{125} x_{3}=\frac{358}{625}$$