Solve {l}{2/25v_{1}+1/50v_{2}+1/20v_{3}=-4}{1/50v_{1}-frac{9}{200}v_{2}+frac{1}{40}v_{3}=-8}{frac{1}{20}v_{1}+frac{1}{40}v_{2}-frac{23}{200}v_{3}=2}

$$\left. \begin{array} { l } { \frac { 2 } { 25 } v _ { 1 } + \frac { 1 } { 50 } v _ { 2 } + \frac { 1 } { 20 } v _ { 3 } = - 4 } \\ { \frac { 1 } { 50 } v _ { 1 } - \frac { 9 } { 200 } v _ { 2 } + \frac { 1 } { 40 } v _ { 3 } = - 8 } \\ { \frac { 1 } { 20 } v _ { 1 } + \frac { 1 } { 40 } v _ { 2 } - \frac { 23 } { 200 } v _ { 3 } = 2 } \end{array} \right.$$

Solve for v_1, v_2, v_3

$$v_{1} = -\frac{16440}{229} = -71\frac{181}{229} \approx -71.790393013$$

$$v_{2} = \frac{30960}{229} = 135\frac{45}{229} \approx 135.19650655$$

$$v_{3} = -\frac{4400}{229} = -19\frac{49}{229} \approx -19.213973799$$

Short Steps Using Substitution

Solve \frac{2}{25}v_{1}+\frac{1}{50}v_{2}+\frac{1}{20}v_{3}=-4 for v_{1}.

$$v_{1}=-50-\frac{1}{4}v_{2}-\frac{5}{8}v_{3}$$

Substitute -50-\frac{1}{4}v_{2}-\frac{5}{8}v_{3} for v_{1} in the second and third equation.

$$\frac{1}{50}\left(-50-\frac{1}{4}v_{2}-\frac{5}{8}v_{3}\right)-\frac{9}{200}v_{2}+\frac{1}{40}v_{3}=-8 \frac{1}{20}\left(-50-\frac{1}{4}v_{2}-\frac{5}{8}v_{3}\right)+\frac{1}{40}v_{2}-\frac{23}{200}v_{3}=2$$

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

$$v_{2}=140+\frac{1}{4}v_{3} v_{3}=-\frac{400}{13}+\frac{10}{117}v_{2}$$

Substitute 140+\frac{1}{4}v_{3} for v_{2} in the equation v_{3}=-\frac{400}{13}+\frac{10}{117}v_{2}.

$$v_{3}=-\frac{400}{13}+\frac{10}{117}\left(140+\frac{1}{4}v_{3}\right)$$

Solve v_{3}=-\frac{400}{13}+\frac{10}{117}\left(140+\frac{1}{4}v_{3}\right) for v_{3}.

$$v_{3}=-\frac{4400}{229}$$

Substitute -\frac{4400}{229} for v_{3} in the equation v_{2}=140+\frac{1}{4}v_{3}.

$$v_{2}=140+\frac{1}{4}\left(-\frac{4400}{229}\right)$$

Calculate v_{2} from v_{2}=140+\frac{1}{4}\left(-\frac{4400}{229}\right).

$$v_{2}=\frac{30960}{229}$$

Substitute \frac{30960}{229} for v_{2} and -\frac{4400}{229} for v_{3} in the equation v_{1}=-50-\frac{1}{4}v_{2}-\frac{5}{8}v_{3}.

$$v_{1}=-50-\frac{1}{4}\times \frac{30960}{229}-\frac{5}{8}\left(-\frac{4400}{229}\right)$$

Calculate v_{1} from v_{1}=-50-\frac{1}{4}\times \frac{30960}{229}-\frac{5}{8}\left(-\frac{4400}{229}\right).

$$v_{1}=-\frac{16440}{229}$$

The system is now solved.

$$v_{1}=-\frac{16440}{229} v_{2}=\frac{30960}{229} v_{3}=-\frac{4400}{229}$$