Solve {l}{frac{10-V_{1}}{10}-frac{V_{1}}{21}-3-frac{V_{1}-V_{2}}{10}=0}{3+frac{V_{1}-V_{2}}{10}-frac{V_{2}+5V_{0}}{10}=0}{V_{1}+V_{0}=10}

$$\left\{ \begin{array} { l } { \frac { 10 - V _ { 1 } } { 10 } - \frac { V _ { 1 } } { 21 } - 3 - \frac { V _ { 1 } - V _ { 2 } } { 10 } = 0 } \\ { 3 + \frac { V _ { 1 } - V _ { 2 } } { 10 } - \frac { V _ { 2 } + 5 V _ { 0 } } { 10 } = 0 } \\ { V _ { 1 } + V _ { 0 } = 10 } \end{array} \right.$$

Solve for V_1, V_2, V_0

$$V_{1} = \frac{630}{11} = 57\frac{3}{11} \approx 57.272727273$$

$$V_{2} = \frac{1780}{11} = 161\frac{9}{11} \approx 161.818181818$$

$$V_{0} = -\frac{520}{11} = -47\frac{3}{11} \approx -47.272727273$$

Short Steps Using Substitution

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

$$-420-52V_{1}+21V_{2}=0 30+V_{1}-2V_{2}-5V_{0}=0 V_{1}+V_{0}=10$$

Reorder the equations.

$$30+V_{1}-2V_{2}-5V_{0}=0 -420-52V_{1}+21V_{2}=0 V_{1}+V_{0}=10$$

Solve 30+V_{1}-2V_{2}-5V_{0}=0 for V_{1}.

$$V_{1}=-30+2V_{2}+5V_{0}$$

Substitute -30+2V_{2}+5V_{0} for V_{1} in the second and third equation.

$$-420-52\left(-30+2V_{2}+5V_{0}\right)+21V_{2}=0 -30+2V_{2}+5V_{0}+V_{0}=10$$

Solve these equations for V_{2} and V_{0} respectively.

$$V_{2}=\frac{1140}{83}-\frac{260}{83}V_{0} V_{0}=\frac{20}{3}-\frac{1}{3}V_{2}$$

Substitute \frac{1140}{83}-\frac{260}{83}V_{0} for V_{2} in the equation V_{0}=\frac{20}{3}-\frac{1}{3}V_{2}.

$$V_{0}=\frac{20}{3}-\frac{1}{3}\left(\frac{1140}{83}-\frac{260}{83}V_{0}\right)$$

Solve V_{0}=\frac{20}{3}-\frac{1}{3}\left(\frac{1140}{83}-\frac{260}{83}V_{0}\right) for V_{0}.

$$V_{0}=-\frac{520}{11}$$

Substitute -\frac{520}{11} for V_{0} in the equation V_{2}=\frac{1140}{83}-\frac{260}{83}V_{0}.

$$V_{2}=\frac{1140}{83}-\frac{260}{83}\left(-\frac{520}{11}\right)$$

Calculate V_{2} from V_{2}=\frac{1140}{83}-\frac{260}{83}\left(-\frac{520}{11}\right).

$$V_{2}=\frac{1780}{11}$$

Substitute \frac{1780}{11} for V_{2} and -\frac{520}{11} for V_{0} in the equation V_{1}=-30+2V_{2}+5V_{0}.

$$V_{1}=-30+2\times \frac{1780}{11}+5\left(-\frac{520}{11}\right)$$

Calculate V_{1} from V_{1}=-30+2\times \frac{1780}{11}+5\left(-\frac{520}{11}\right).

$$V_{1}=\frac{630}{11}$$

The system is now solved.

$$V_{1}=\frac{630}{11} V_{2}=\frac{1780}{11} V_{0}=-\frac{520}{11}$$