User:Tedburke/sandbox

$$I_1 = I_2 + I_3$$

There are three terms in this equation: $$I_1$$, $$I_2$$ and $$I_3$$. Replace the first and third terms with expressions for $$I_1$$ and $$I_3$$ in terms of $$I_2$$...

$$\frac{10-I_2R_2}{R_1} = I_2 + \frac{I_2R_2}{R_3+R_4}$$

Now, in order to eliminate the fractions from the equation, multiply each of the three terms by $$R_1(R_3 + R_4)$$.

$$\frac{(10-I_2R_2)\cancel{R_1}(R_3+R_4)}{\cancel{R_1}} = I_2R_1(R_3+R_4) + \frac{I_2R_2R_1\cancel{(R_3+R_4)}}{\cancel{R_3+R_4}}$$

Removing the terms that cancel...

$$(10-I_2R_2)(R_3+R_4) = I_2R_1(R_3+R_4) + I_2R_2R_1$$

Multiplying out the left side of the equation...

$$10(R_3+R_4)-I_2R_2(R_3+R_4) = I_2R_1(R_3+R_4) + I_2R_2R_1$$

Moving the second term from the left hand side to the right hand side...

$$10(R_3+R_4) = I_2R_1(R_3+R_4) + I_2R_2R_1 + I_2R_2(R_3+R_4)$$

Now, $$I_2$$ is a factor in all three terms on the left hand side, so we can break it out as follows...

$$10(R_3+R_4) = I_2(R_1(R_3+R_4) + R_2R_1 + R_2(R_3+R_4))$$

Dividing both sides by all that stuff in the brackets on the right hand side...

$$\frac{10(R_3+R_4)}{R_1(R_3+R_4) + R_2R_1 + R_2(R_3+R_4)} = I_2$$

Swap the left and right sides and tidy up the denominator of the fraction a bit...

$$I_2 = \frac{10(R_3+R_4)}{R_2R_1 + (R_1+R_2)(R_3+R_4)}$$