Solve {l}{4I_{1}-4I_{2}=7}{-4I_{1}+28I_{2}-10I_{3}=0}{-10I_{2}+18I_{3}=3}

$$\left. \begin{array} { l } { 4 I _ { 1 } - 4 I _ { 2 } = 7 } \\ { - 4 I _ { 1 } + 28 I _ { 2 } - 10 I _ { 3 } = 0 } \\ { - 10 I _ { 2 } + 18 I _ { 3 } = 3 } \end{array} \right.$$

Solve for I_1, I_2, I_3

$$I_{1} = \frac{737}{332} = 2\frac{73}{332} \approx 2.219879518$$

$$I_{2}=\frac{39}{83}\approx 0.469879518$$

$$I_{3}=\frac{71}{166}\approx 0.427710843$$

Short Steps Using Substitution

Solve 4I_{1}-4I_{2}=7 for I_{1}.

$$I_{1}=I_{2}+\frac{7}{4}$$

Substitute I_{2}+\frac{7}{4} for I_{1} in the equation -4I_{1}+28I_{2}-10I_{3}=0.

$$-4\left(I_{2}+\frac{7}{4}\right)+28I_{2}-10I_{3}=0$$

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

$$I_{2}=\frac{7}{24}+\frac{5}{12}I_{3} I_{3}=\frac{5}{9}I_{2}+\frac{1}{6}$$

Substitute \frac{7}{24}+\frac{5}{12}I_{3} for I_{2} in the equation I_{3}=\frac{5}{9}I_{2}+\frac{1}{6}.

$$I_{3}=\frac{5}{9}\left(\frac{7}{24}+\frac{5}{12}I_{3}\right)+\frac{1}{6}$$

Solve I_{3}=\frac{5}{9}\left(\frac{7}{24}+\frac{5}{12}I_{3}\right)+\frac{1}{6} for I_{3}.

$$I_{3}=\frac{71}{166}$$

Substitute \frac{71}{166} for I_{3} in the equation I_{2}=\frac{7}{24}+\frac{5}{12}I_{3}.

$$I_{2}=\frac{7}{24}+\frac{5}{12}\times \frac{71}{166}$$

Calculate I_{2} from I_{2}=\frac{7}{24}+\frac{5}{12}\times \frac{71}{166}.

$$I_{2}=\frac{39}{83}$$

Substitute \frac{39}{83} for I_{2} in the equation I_{1}=I_{2}+\frac{7}{4}.

$$I_{1}=\frac{39}{83}+\frac{7}{4}$$

Calculate I_{1} from I_{1}=\frac{39}{83}+\frac{7}{4}.

$$I_{1}=\frac{737}{332}$$

The system is now solved.

$$I_{1}=\frac{737}{332} I_{2}=\frac{39}{83} I_{3}=\frac{71}{166}$$