Solve {l}{I_{1}+I_{2}+I_{3}=0}{10I_{1}-7,5I_{2}=130-125}{7,5I_{2}-10I_{3}=125-50}

$$\left\{ \begin{array} { l } { I _ { 1 } + I _ { 2 } + I _ { 3 } = 0 } \\ { 10 I _ { 1 } - 7,5 I _ { 2 } = 130 - 125 } \\ { 7,5 I _ { 2 } - 10 I _ { 3 } = 125 - 50 } \end{array} \right.$$

Solve for I_1, I_2, I_3

$$I_{1}=2,6$$

$$I_{2}=2,8$$

$$I_{3}=-5,4$$

Short Steps Using Substitution

Solve I_{1}+I_{2}+I_{3}=0 for I_{1}.

$$I_{1}=-I_{2}-I_{3}$$

Substitute -I_{2}-I_{3} for I_{1} in the equation 10I_{1}-7,5I_{2}=130-125.

$$10\left(-I_{2}-I_{3}\right)-7,5I_{2}=130-125$$

Solve the second equation for I_{2} and the third equation for I_{3}.

$$I_{2}=-\frac{2}{7}-\frac{4}{7}I_{3} I_{3}=-\frac{15}{2}+\frac{3}{4}I_{2}$$

Substitute -\frac{2}{7}-\frac{4}{7}I_{3} for I_{2} in the equation I_{3}=-\frac{15}{2}+\frac{3}{4}I_{2}.

$$I_{3}=-\frac{15}{2}+\frac{3}{4}\left(-\frac{2}{7}-\frac{4}{7}I_{3}\right)$$

Solve I_{3}=-\frac{15}{2}+\frac{3}{4}\left(-\frac{2}{7}-\frac{4}{7}I_{3}\right) for I_{3}.

$$I_{3}=-\frac{27}{5}$$

Substitute -\frac{27}{5} for I_{3} in the equation I_{2}=-\frac{2}{7}-\frac{4}{7}I_{3}.

$$I_{2}=-\frac{2}{7}-\frac{4}{7}\left(-\frac{27}{5}\right)$$

Calculate I_{2} from I_{2}=-\frac{2}{7}-\frac{4}{7}\left(-\frac{27}{5}\right).

$$I_{2}=\frac{14}{5}$$

Substitute \frac{14}{5} for I_{2} and -\frac{27}{5} for I_{3} in the equation I_{1}=-I_{2}-I_{3}.

$$I_{1}=-\frac{14}{5}-\left(-\frac{27}{5}\right)$$

Calculate I_{1} from I_{1}=-\frac{14}{5}-\left(-\frac{27}{5}\right).

$$I_{1}=\frac{13}{5}$$

The system is now solved.

$$I_{1}=\frac{13}{5} I_{2}=\frac{14}{5} I_{3}=-\frac{27}{5}$$