Solve {l}{1/10U_{1}+(U_{1}-U_{2})1/2=5I_{x}+2}{1/2(U_{1}-U_{2})+5I_{x}+1/4U_{2}=0}{I_{x}=frac{6}{13}}

$$\left. \begin{array} { l } { \frac { 1 } { 10 } U _ { 1 } + ( U _ { 1 } - U _ { 2 } ) \frac { 1 } { 2 } = 5 I _ { x } + 2 } \\ { \frac { 1 } { 2 } ( U _ { 1 } - U _ { 2 } ) + 5 I _ { x } + \frac { 1 } { 4 } U _ { 2 } = 0 } \\ { I _ { x } = \frac { 6 } { 13 } } \end{array} \right.$$

Solve for U_1, U_2, I_x

$$U_{1} = -\frac{290}{13} = -22\frac{4}{13} \approx -22.307692308$$

$$U_{2} = -\frac{460}{13} = -35\frac{5}{13} \approx -35.384615385$$

$$I_{x}=\frac{6}{13}\approx 0.461538462$$

Short Steps Using Substitution

Reorder the equations.

$$I_{x}=\frac{6}{13} \frac{1}{2}\left(U_{1}-U_{2}\right)+5I_{x}+\frac{1}{4}U_{2}=0 \frac{1}{10}U_{1}+\left(U_{1}-U_{2}\right)\times \frac{1}{2}=5I_{x}+2$$

Substitute \frac{6}{13} for I_{x} in the second and third equation.

$$\frac{1}{2}\left(U_{1}-U_{2}\right)+5\times \frac{6}{13}+\frac{1}{4}U_{2}=0 \frac{1}{10}U_{1}+\left(U_{1}-U_{2}\right)\times \frac{1}{2}=5\times \frac{6}{13}+2$$

Solve these equations for U_{2} and U_{1} respectively.

$$U_{2}=\frac{120}{13}+2U_{1} U_{1}=\frac{280}{39}+\frac{5}{6}U_{2}$$

Substitute \frac{120}{13}+2U_{1} for U_{2} in the equation U_{1}=\frac{280}{39}+\frac{5}{6}U_{2}.

$$U_{1}=\frac{280}{39}+\frac{5}{6}\left(\frac{120}{13}+2U_{1}\right)$$

Solve U_{1}=\frac{280}{39}+\frac{5}{6}\left(\frac{120}{13}+2U_{1}\right) for U_{1}.

$$U_{1}=-\frac{290}{13}$$

Substitute -\frac{290}{13} for U_{1} in the equation U_{2}=\frac{120}{13}+2U_{1}.

$$U_{2}=\frac{120}{13}+2\left(-\frac{290}{13}\right)$$

Calculate U_{2} from U_{2}=\frac{120}{13}+2\left(-\frac{290}{13}\right).

$$U_{2}=-\frac{460}{13}$$

The system is now solved.

$$U_{1}=-\frac{290}{13} U_{2}=-\frac{460}{13} I_{x}=\frac{6}{13}$$