Solve {l}{21=5I_{1}+6I_{2}}{14=10I_{3}+6I_{2}}{I_{1}=I_{2}-I_{3}}

$$\left. \begin{array}{l}{ 21 = 5 I _ { 1 } + 6 I _ { 2 } }\\{ 14 = 10 I _ { 3 } + 6 I _ { 2 } }\\{ I _ { 1 } = I _ { 2 } - I _ { 3 } }\end{array} \right.$$

Solve for I_1, I_2, I_3

$$I_{1} = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

$$I_{2}=2$$

$$I_{3}=\frac{1}{5}=0.2$$

Short Steps Using Substitution

Reorder the equations.

$$I_{1}=I_{2}-I_{3} 14=10I_{3}+6I_{2} 21=5I_{1}+6I_{2}$$

Substitute I_{2}-I_{3} for I_{1} in the equation 21=5I_{1}+6I_{2}.

$$21=5\left(I_{2}-I_{3}\right)+6I_{2}$$

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

$$I_{2}=\frac{7}{3}-\frac{5}{3}I_{3} I_{3}=\frac{11}{5}I_{2}-\frac{21}{5}$$

Substitute \frac{7}{3}-\frac{5}{3}I_{3} for I_{2} in the equation I_{3}=\frac{11}{5}I_{2}-\frac{21}{5}.

$$I_{3}=\frac{11}{5}\left(\frac{7}{3}-\frac{5}{3}I_{3}\right)-\frac{21}{5}$$

Solve I_{3}=\frac{11}{5}\left(\frac{7}{3}-\frac{5}{3}I_{3}\right)-\frac{21}{5} for I_{3}.

$$I_{3}=\frac{1}{5}$$

Substitute \frac{1}{5} for I_{3} in the equation I_{2}=\frac{7}{3}-\frac{5}{3}I_{3}.

$$I_{2}=\frac{7}{3}-\frac{5}{3}\times \frac{1}{5}$$

Calculate I_{2} from I_{2}=\frac{7}{3}-\frac{5}{3}\times \frac{1}{5}.

$$I_{2}=2$$

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

$$I_{1}=2-\frac{1}{5}$$

Calculate I_{1} from I_{1}=2-\frac{1}{5}.

$$I_{1}=\frac{9}{5}$$

The system is now solved.

$$I_{1}=\frac{9}{5} I_{2}=2 I_{3}=\frac{1}{5}$$