Solve {l}{I_{1}+I_{2}=I}{4I_{1}-2I_{2}=20-10}{2I_{2}+2I=10}

$$\left. \begin{array} { l } { I _ { 1 } + I _ { 2 } = I } \\ { 4 I _ { 1 } - 2 I _ { 2 } = 20 - 10 } \\ { 2 I _ { 2 } + 2 I = 10 } \end{array} \right.$$

Solve for I_1, I_2, I

$$I_{1}=3$$

$$I_{2}=1$$

$$I=4$$

Short Steps Using Substitution

Solve I_{1}+I_{2}=I for I.

$$I=I_{1}+I_{2}$$

Substitute I_{1}+I_{2} for I in the equation 2I_{2}+2I=10.

$$2I_{2}+2\left(I_{1}+I_{2}\right)=10$$

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

$$I_{2}=-5+2I_{1} I_{1}=-2I_{2}+5$$

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

$$I_{1}=-2\left(-5+2I_{1}\right)+5$$

Solve I_{1}=-2\left(-5+2I_{1}\right)+5 for I_{1}.

$$I_{1}=3$$

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

$$I_{2}=-5+2\times 3$$

Calculate I_{2} from I_{2}=-5+2\times 3.

$$I_{2}=1$$

Substitute 1 for I_{2} and 3 for I_{1} in the equation I=I_{1}+I_{2}.

$$I=3+1$$

Calculate I from I=3+1.

$$I=4$$

The system is now solved.

$$I_{1}=3 I_{2}=1 I=4$$