Solve {l}{I_{1}-I_{2}-I_{3}=0}{100I_{1}-1168I_{3}=74,06}{-I_{3}*(1168)+155I_{2}=14,06}

$$\left\{ \begin{array} { l } { I _ { 1 } - I _ { 2 } - I _ { 3 } = 0 } \\ { 100 I _ { 1 } - 1168 I _ { 3 } = 74,06 } \\ { - I _ { 3 } \cdot ( 1168 ) + 155 I _ { 2 } = 14,06 } \end{array} \right.$$

Solve for I_1, I_2, I_3

$$I_{1}=-\frac{815593}{487400}\approx -1,673354534$$

$$I_{2}=-\frac{35743}{24370}\approx -1,466680345$$

$$I_{3}=-\frac{100733}{487400}\approx -0,20667419$$

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 100I_{1}-1168I_{3}=74,06.

$$100\left(I_{2}+I_{3}\right)-1168I_{3}=74,06$$

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

$$I_{2}=\frac{267}{25}I_{3}+\frac{3703}{5000} I_{3}=\frac{155}{1168}I_{2}-\frac{703}{58400}$$

Substitute \frac{267}{25}I_{3}+\frac{3703}{5000} for I_{2} in the equation I_{3}=\frac{155}{1168}I_{2}-\frac{703}{58400}.

$$I_{3}=\frac{155}{1168}\left(\frac{267}{25}I_{3}+\frac{3703}{5000}\right)-\frac{703}{58400}$$

Solve I_{3}=\frac{155}{1168}\left(\frac{267}{25}I_{3}+\frac{3703}{5000}\right)-\frac{703}{58400} for I_{3}.

$$I_{3}=-\frac{100733}{487400}$$

Substitute -\frac{100733}{487400} for I_{3} in the equation I_{2}=\frac{267}{25}I_{3}+\frac{3703}{5000}.

$$I_{2}=\frac{267}{25}\left(-\frac{100733}{487400}\right)+\frac{3703}{5000}$$

Calculate I_{2} from I_{2}=\frac{267}{25}\left(-\frac{100733}{487400}\right)+\frac{3703}{5000}.

$$I_{2}=-\frac{35743}{24370}$$

Substitute -\frac{35743}{24370} for I_{2} and -\frac{100733}{487400} for I_{3} in the equation I_{1}=I_{2}+I_{3}.

$$I_{1}=-\frac{35743}{24370}-\frac{100733}{487400}$$

Calculate I_{1} from I_{1}=-\frac{35743}{24370}-\frac{100733}{487400}.

$$I_{1}=-\frac{815593}{487400}$$

The system is now solved.

$$I_{1}=-\frac{815593}{487400} I_{2}=-\frac{35743}{24370} I_{3}=-\frac{100733}{487400}$$