Solve {l}{5v_{1}+2v_{2}-2v_{3}=0}{v_{1}-v_{2}=25}{frac{v_{1}}{2}+frac{v_{2}}{4}+frac{v_{3}}{3}=0}

$$\left. \begin{array} { l } { 5 v _ { 1 } + 2 v _ { 2 } - 2 v _ { 3 } = 0 } \\ { v _ { 1 } - v _ { 2 } = 25 } \\ { \frac { v _ { 1 } } { 2 } + \frac { v _ { 2 } } { 4 } + \frac { v _ { 3 } } { 3 } = 0 } \end{array} \right.$$

Solve for v_1, v_2, v_3

$$v_{1} = \frac{175}{23} = 7\frac{14}{23} \approx 7.608695652$$

$$v_{2} = -\frac{400}{23} = -17\frac{9}{23} \approx -17.391304348$$

$$v_{3} = \frac{75}{46} = 1\frac{29}{46} \approx 1.630434783$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$5v_{1}+2v_{2}-2v_{3}=0 v_{1}-v_{2}=25 6v_{1}+3v_{2}+4v_{3}=0$$

Reorder the equations.

$$v_{1}-v_{2}=25 5v_{1}+2v_{2}-2v_{3}=0 6v_{1}+3v_{2}+4v_{3}=0$$

Solve v_{1}-v_{2}=25 for v_{1}.

$$v_{1}=v_{2}+25$$

Substitute v_{2}+25 for v_{1} in the second and third equation.

$$5\left(v_{2}+25\right)+2v_{2}-2v_{3}=0 6\left(v_{2}+25\right)+3v_{2}+4v_{3}=0$$

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

$$v_{2}=-\frac{125}{7}+\frac{2}{7}v_{3} v_{3}=-\frac{75}{2}-\frac{9}{4}v_{2}$$

Substitute -\frac{125}{7}+\frac{2}{7}v_{3} for v_{2} in the equation v_{3}=-\frac{75}{2}-\frac{9}{4}v_{2}.

$$v_{3}=-\frac{75}{2}-\frac{9}{4}\left(-\frac{125}{7}+\frac{2}{7}v_{3}\right)$$

Solve v_{3}=-\frac{75}{2}-\frac{9}{4}\left(-\frac{125}{7}+\frac{2}{7}v_{3}\right) for v_{3}.

$$v_{3}=\frac{75}{46}$$

Substitute \frac{75}{46} for v_{3} in the equation v_{2}=-\frac{125}{7}+\frac{2}{7}v_{3}.

$$v_{2}=-\frac{125}{7}+\frac{2}{7}\times \frac{75}{46}$$

Calculate v_{2} from v_{2}=-\frac{125}{7}+\frac{2}{7}\times \frac{75}{46}.

$$v_{2}=-\frac{400}{23}$$

Substitute -\frac{400}{23} for v_{2} in the equation v_{1}=v_{2}+25.

$$v_{1}=-\frac{400}{23}+25$$

Calculate v_{1} from v_{1}=-\frac{400}{23}+25.

$$v_{1}=\frac{175}{23}$$

The system is now solved.

$$v_{1}=\frac{175}{23} v_{2}=-\frac{400}{23} v_{3}=\frac{75}{46}$$