Solve {l}{4I_{1}-4I_{2}=7}{-4I_{1}+28I_{2}-10I_{3}=0}{-10I_{2}+1813=3}

$$\left. \begin{array} { l } { 4 I _ { 1 } - 4 I _ { 2 } = 7 } \\ { - 4 I _ { 1 } + 28 I _ { 2 } - 10 I _ { 3 } = 0 } \\ { - 10 I _ { 2 } + 1813 = 3 } \end{array} \right.$$

Solve for I_1, I_2, I_3

$$I_{1} = \frac{731}{4} = 182\frac{3}{4} = 182.75$$

$$I_{2}=181$$

$$I_{3} = \frac{4337}{10} = 433\frac{7}{10} = 433.7$$

Solution Steps

Consider the third equation. Subtract 1813 from both sides.

$$-10I_{2}=3-1813$$

Subtract 1813 from 3 to get -1810.

$$-10I_{2}=-1810$$

Divide both sides by -10.

$$I_{2}=\frac{-1810}{-10}$$

Divide -1810 by -10 to get 181.

$$I_{2}=181$$

Consider the first equation. Insert the known values of variables into the equation.

$$4I_{1}-4\times 181=7$$

Multiply -4 and 181 to get -724.

$$4I_{1}-724=7$$

Add 724 to both sides.

$$4I_{1}=7+724$$

Add 7 and 724 to get 731.

$$4I_{1}=731$$

Divide both sides by 4.

$$I_{1}=\frac{731}{4}$$

Consider the second equation. Insert the known values of variables into the equation.

$$-4\times \frac{731}{4}+28\times 181-10I_{3}=0$$

Multiply -4 and \frac{731}{4} to get -731.

$$-731+28\times 181-10I_{3}=0$$

Multiply 28 and 181 to get 5068.

$$-731+5068-10I_{3}=0$$

Add -731 and 5068 to get 4337.

$$4337-10I_{3}=0$$

Subtract 4337 from both sides. Anything subtracted from zero gives its negation.

$$-10I_{3}=-4337$$

Divide both sides by -10.

$$I_{3}=\frac{-4337}{-10}$$

Fraction \frac{-4337}{-10} can be simplified to \frac{4337}{10} by removing the negative sign from both the numerator and the denominator.

$$I_{3}=\frac{4337}{10}$$

The system is now solved.

$$I_{1}=\frac{731}{4} I_{2}=181 I_{3}=\frac{4337}{10}$$