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

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

Solve for U_1, U_2, I_x

$$U_{1} = -\frac{20}{9} = -2\frac{2}{9} \approx -2.222222222$$

$$U_{2} = -\frac{80}{9} = -8\frac{8}{9} \approx -8.888888889$$

$$I_{x}=-\frac{2}{9}\approx -0.222222222$$

Short Steps Using Substitution

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

$$6U_{1}-5U_{2}+50I_{x}=20 -U_{2}+2U_{1}+20I_{x}=0 10I_{x}=U_{1}$$

Reorder the equations.

$$10I_{x}=U_{1} -U_{2}+2U_{1}+20I_{x}=0 6U_{1}-5U_{2}+50I_{x}=20$$

Solve 10I_{x}=U_{1} for U_{1}.

$$U_{1}=10I_{x}$$

Substitute 10I_{x} for U_{1} in the second and third equation.

$$-U_{2}+2\times 10I_{x}+20I_{x}=0 6\times 10I_{x}-5U_{2}+50I_{x}=20$$

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

$$U_{2}=40I_{x} I_{x}=\frac{2}{11}+\frac{1}{22}U_{2}$$

Substitute 40I_{x} for U_{2} in the equation I_{x}=\frac{2}{11}+\frac{1}{22}U_{2}.

$$I_{x}=\frac{2}{11}+\frac{1}{22}\times 40I_{x}$$

Solve I_{x}=\frac{2}{11}+\frac{1}{22}\times 40I_{x} for I_{x}.

$$I_{x}=-\frac{2}{9}$$

Substitute -\frac{2}{9} for I_{x} in the equation U_{2}=40I_{x}.

$$U_{2}=40\left(-\frac{2}{9}\right)$$

Calculate U_{2} from U_{2}=40\left(-\frac{2}{9}\right).

$$U_{2}=-\frac{80}{9}$$

Substitute -\frac{2}{9} for I_{x} in the equation U_{1}=10I_{x}.

$$U_{1}=10\left(-\frac{2}{9}\right)$$

Calculate U_{1} from U_{1}=10\left(-\frac{2}{9}\right).

$$U_{1}=-\frac{20}{9}$$

The system is now solved.

$$U_{1}=-\frac{20}{9} U_{2}=-\frac{80}{9} I_{x}=-\frac{2}{9}$$