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

$$v_{1} = \frac{16440}{119} = 138\frac{18}{119} \approx 138.151260504$$

$$v_{2} = \frac{18960}{119} = 159\frac{39}{119} \approx 159.327731092$$

$$v_{3} = \frac{9200}{119} = 77\frac{37}{119} \approx 77.31092437$$

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.

$$v_{2}=140+\frac{1}{4}v_{3} v_{3}=\frac{400}{67}+\frac{30}{67}v_{2}$$

Substitute 140+\frac{1}{4}v_{3} for v_{2} in the equation v_{3}=\frac{400}{67}+\frac{30}{67}v_{2}.

$$v_{3}=\frac{400}{67}+\frac{30}{67}\left(140+\frac{1}{4}v_{3}\right)$$

Solve v_{3}=\frac{400}{67}+\frac{30}{67}\left(140+\frac{1}{4}v_{3}\right) for v_{3}.

$$v_{3}=\frac{9200}{119}$$

Substitute \frac{9200}{119} for v_{3} in the equation v_{2}=140+\frac{1}{4}v_{3}.

$$v_{2}=140+\frac{1}{4}\times \frac{9200}{119}$$

Calculate v_{2} from v_{2}=140+\frac{1}{4}\times \frac{9200}{119}.

$$v_{2}=\frac{18960}{119}$$

Substitute \frac{18960}{119} for v_{2} and \frac{9200}{119} for v_{3} in the equation v_{1}=50+\frac{1}{4}v_{2}+\frac{5}{8}v_{3}.

$$v_{1}=50+\frac{1}{4}\times \frac{18960}{119}+\frac{5}{8}\times \frac{9200}{119}$$

Calculate v_{1} from v_{1}=50+\frac{1}{4}\times \frac{18960}{119}+\frac{5}{8}\times \frac{9200}{119}.

$$v_{1}=\frac{16440}{119}$$

The system is now solved.

$$v_{1}=\frac{16440}{119} v_{2}=\frac{18960}{119} v_{3}=\frac{9200}{119}$$