Solve {l}{36=I_{1}(9+4)-I_{2}(-9-4)-[3(0)}{0=I_{1}(-9-4)+I_{2}(4+9+6)-I_{3}(-6)}{0=I_{1}(0)-I_{2}(-6)+I_{3}(6+3+6)}

$$\left. \begin{array} { l } { 36 = I _ { 1 } ( 9 + 4 ) - I _ { 2 } ( - 9 - 4 ) - [ 3 ( 0 ) } \\ { 0 = I _ { 1 } ( - 9 - 4 ) + I _ { 2 } ( 4 + 9 + 6 ) - I _ { 3 } ( - 6 ) } \\ { 0 = I _ { 1 } ( 0 ) - I _ { 2 } ( - 6 ) + I _ { 3 } ( 6 + 3 + 6 ) } \end{array} \right.$$

Solve for I_1, I_2, I_3

$$I_{1} = \frac{747}{481} = 1\frac{266}{481} \approx 1.553014553$$

$$I_{2} = \frac{45}{37} = 1\frac{8}{37} \approx 1.216216216$$

$$I_{3}=-\frac{18}{37}\approx -0.486486486$$

Short Steps Using Substitution

Solve 36=I_{1}\left(9+4\right)-I_{2}\left(-9-4\right)-3\times 0 for I_{1}.

$$I_{1}=-I_{2}+\frac{36}{13}$$

Substitute -I_{2}+\frac{36}{13} for I_{1} in the second and third equation.

$$0=\left(-I_{2}+\frac{36}{13}\right)\left(-9-4\right)+I_{2}\left(4+9+6\right)-I_{3}\left(-6\right) 0=\left(-I_{2}+\frac{36}{13}\right)\times 0-I_{2}\left(-6\right)+I_{3}\left(6+3+6\right)$$

Solve these equations for I_{2} and I_{3} respectively.

$$I_{2}=\frac{9}{8}-\frac{3}{16}I_{3} I_{3}=-\frac{2}{5}I_{2}$$

Substitute \frac{9}{8}-\frac{3}{16}I_{3} for I_{2} in the equation I_{3}=-\frac{2}{5}I_{2}.

$$I_{3}=-\frac{2}{5}\left(\frac{9}{8}-\frac{3}{16}I_{3}\right)$$

Solve I_{3}=-\frac{2}{5}\left(\frac{9}{8}-\frac{3}{16}I_{3}\right) for I_{3}.

$$I_{3}=-\frac{18}{37}$$

Substitute -\frac{18}{37} for I_{3} in the equation I_{2}=\frac{9}{8}-\frac{3}{16}I_{3}.

$$I_{2}=\frac{9}{8}-\frac{3}{16}\left(-\frac{18}{37}\right)$$

Calculate I_{2} from I_{2}=\frac{9}{8}-\frac{3}{16}\left(-\frac{18}{37}\right).

$$I_{2}=\frac{45}{37}$$

Substitute \frac{45}{37} for I_{2} in the equation I_{1}=-I_{2}+\frac{36}{13}.

$$I_{1}=-\frac{45}{37}+\frac{36}{13}$$

Calculate I_{1} from I_{1}=-\frac{45}{37}+\frac{36}{13}.

$$I_{1}=\frac{747}{481}$$

The system is now solved.

$$I_{1}=\frac{747}{481} I_{2}=\frac{45}{37} I_{3}=-\frac{18}{37}$$