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}{U_{1}=4.6}

$$\left. \begin{array} { l } { \frac { 1 } { 10 } U _ { 1 } + ( U _ { 1 } - U _ { 2 } ) \frac { 1 } { 2 } = 5 I _ { x } + 2 } \\ { \frac { 1 } { 2 } ( U _ { 1 } - U _ { 2 } ) + 5 I _ { x } + \frac { 1 } { 4 } U _ { 2 } = 0 } \\ { U _ { 1 } = 4.6 } \end{array} \right.$$

Solve for U_1, U_2, I_x

$$U_{1}=4.6$$

$$U_{2}=4.08$$

$$I_{x}=-0.256$$

Short Steps Using Substitution

Reorder the equations.

$$U_{1}=4.6 \frac{1}{2}\left(U_{1}-U_{2}\right)+5I_{x}+\frac{1}{4}U_{2}=0 \frac{1}{10}U_{1}+\left(U_{1}-U_{2}\right)\times \frac{1}{2}=5I_{x}+2$$

Substitute 4.6 for U_{1} in the second and third equation.

$$\frac{1}{2}\left(4.6-U_{2}\right)+5I_{x}+\frac{1}{4}U_{2}=0 \frac{1}{10}\times 4.6+\left(4.6-U_{2}\right)\times \frac{1}{2}=5I_{x}+2$$

Solve these equations for U_{2} and I_{x} respectively.

$$U_{2}=\frac{46}{5}+20I_{x} I_{x}=\frac{19}{125}-\frac{1}{10}U_{2}$$

Substitute \frac{46}{5}+20I_{x} for U_{2} in the equation I_{x}=\frac{19}{125}-\frac{1}{10}U_{2}.

$$I_{x}=\frac{19}{125}-\frac{1}{10}\left(\frac{46}{5}+20I_{x}\right)$$

Solve I_{x}=\frac{19}{125}-\frac{1}{10}\left(\frac{46}{5}+20I_{x}\right) for I_{x}.

$$I_{x}=-\frac{32}{125}$$

Substitute -\frac{32}{125} for I_{x} in the equation U_{2}=\frac{46}{5}+20I_{x}.

$$U_{2}=\frac{46}{5}+20\left(-\frac{32}{125}\right)$$

Calculate U_{2} from U_{2}=\frac{46}{5}+20\left(-\frac{32}{125}\right).

$$U_{2}=\frac{102}{25}$$

The system is now solved.

$$U_{1}=4.6 U_{2}=\frac{102}{25} I_{x}=-\frac{32}{125}$$