Solve {l}{V_{A}+V_{B}+V_{C}=40}{4V_{C}+7V_{A}=-188}{4V_{B}+11V_{A}=-28}

$$\left. \begin{array} { l } { V _ { A } + V _ { B } + V _ { C } = 40 } \\ { 4 V _ { C } + 7 V _ { A } = - 188 } \\ { 4 V _ { B } + 11 V _ { A } = - 28 } \end{array} \right.$$

Solve for V_A, V_B, V_C

$$V_{A} = -\frac{188}{7} = -26\frac{6}{7} \approx -26.857142857$$

$$V_{B} = \frac{468}{7} = 66\frac{6}{7} \approx 66.857142857$$

$$V_{C}=0$$

Short Steps Using Substitution

Solve V_{A}+V_{B}+V_{C}=40 for V_{A}.

$$V_{A}=-V_{B}-V_{C}+40$$

Substitute -V_{B}-V_{C}+40 for V_{A} in the second and third equation.

$$4V_{C}+7\left(-V_{B}-V_{C}+40\right)=-188 4V_{B}+11\left(-V_{B}-V_{C}+40\right)=-28$$

Solve these equations for V_{B} and V_{C} respectively.

$$V_{B}=\frac{468}{7}-\frac{3}{7}V_{C} V_{C}=\frac{468}{11}-\frac{7}{11}V_{B}$$

Substitute \frac{468}{7}-\frac{3}{7}V_{C} for V_{B} in the equation V_{C}=\frac{468}{11}-\frac{7}{11}V_{B}.

$$V_{C}=\frac{468}{11}-\frac{7}{11}\left(\frac{468}{7}-\frac{3}{7}V_{C}\right)$$

Solve V_{C}=\frac{468}{11}-\frac{7}{11}\left(\frac{468}{7}-\frac{3}{7}V_{C}\right) for V_{C}.

$$V_{C}=0$$

Substitute 0 for V_{C} in the equation V_{B}=\frac{468}{7}-\frac{3}{7}V_{C}.

$$V_{B}=\frac{468}{7}-\frac{3}{7}\times 0$$

Calculate V_{B} from V_{B}=\frac{468}{7}-\frac{3}{7}\times 0.

$$V_{B}=\frac{468}{7}$$

Substitute \frac{468}{7} for V_{B} and 0 for V_{C} in the equation V_{A}=-V_{B}-V_{C}+40.

$$V_{A}=-\frac{468}{7}-0+40$$

Calculate V_{A} from V_{A}=-\frac{468}{7}-0+40.

$$V_{A}=-\frac{188}{7}$$

The system is now solved.

$$V_{A}=-\frac{188}{7} V_{B}=\frac{468}{7} V_{C}=0$$