Solve {l}{-I_{x}+I_{2}+I_{1}=4}{-4I_{x}=40I_{1}}{40I_{1}=12I_{2}}

$$\left. \begin{array}{l}{ - I _ { x } + I _ { 2 } + I _ { 1 } = 4 }\\{ - 4 I _ { x } = 40 I _ { 1 } }\\{ 40 I _ { 1 } = 12 I _ { 2 } }\end{array} \right.$$

Solve for I_x, I_2, I_1

$$I_{x} = -\frac{120}{43} = -2\frac{34}{43} \approx -2.790697674$$

$$I_{2}=\frac{40}{43}\approx 0.930232558$$

$$I_{1}=\frac{12}{43}\approx 0.279069767$$

Short Steps Using Substitution

Solve -I_{x}+I_{2}+I_{1}=4 for I_{x}.

$$I_{x}=I_{2}+I_{1}-4$$

Substitute I_{2}+I_{1}-4 for I_{x} in the equation -4I_{x}=40I_{1}.

$$-4\left(I_{2}+I_{1}-4\right)=40I_{1}$$

Solve the second equation for I_{2} and the third equation for I_{1}.

$$I_{2}=4-11I_{1} I_{1}=\frac{3}{10}I_{2}$$

Substitute 4-11I_{1} for I_{2} in the equation I_{1}=\frac{3}{10}I_{2}.

$$I_{1}=\frac{3}{10}\left(4-11I_{1}\right)$$

Solve I_{1}=\frac{3}{10}\left(4-11I_{1}\right) for I_{1}.

$$I_{1}=\frac{12}{43}$$

Substitute \frac{12}{43} for I_{1} in the equation I_{2}=4-11I_{1}.

$$I_{2}=4-11\times \frac{12}{43}$$

Calculate I_{2} from I_{2}=4-11\times \frac{12}{43}.

$$I_{2}=\frac{40}{43}$$

Substitute \frac{40}{43} for I_{2} and \frac{12}{43} for I_{1} in the equation I_{x}=I_{2}+I_{1}-4.

$$I_{x}=\frac{40}{43}+\frac{12}{43}-4$$

Calculate I_{x} from I_{x}=\frac{40}{43}+\frac{12}{43}-4.

$$I_{x}=-\frac{120}{43}$$

The system is now solved.

$$I_{x}=-\frac{120}{43} I_{2}=\frac{40}{43} I_{1}=\frac{12}{43}$$