Equation simplification Sorry if the question is pretty trivial but can anyone tell me what rule or law was used to go from $$e_2=C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4-R_{7}R_2^{-1}e_2$$ $$to$$ $$e_2=\dfrac{R_2*C_4^{-1}}{R_2+R_7}x_1+\dfrac{R_2*C_{10}^{-1}}{R_2+R_7}x_2-\dfrac{R_2*R_{7}*I_3^{-1}}{R_2+R_7}x_4$$
 A: $e_2=C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4-R_{7}R_2^{-1}e_2$
$e_2+R_{7}R_2^{-1}e_2 = C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4$
$e_2(1 + R_{7}R_2^{-1})= C_4^{-1}x_1+C_{10}^{-1}x_2-R_{7}I_3^{-1}x_4$
$e_2 = \frac{C_4^{-1}x_1}{1 + R_{7}R_2^{-1}}+\frac{C_{10}^{-1}x_2}{1 + R_{7}R_2^{-1}}-\frac{R_{7}I_3^{-1}x_4}{1 + R_{7}R_2^{-1}}$ 
Multiply each numerator and denominator by $R_2$ to get......... 
$e_2=\dfrac{R_2*C_4^{-1}}{R_2+R_7}x_1+\dfrac{R_2*C_{10}^{-1}}{R_2+R_7}x_2-\dfrac{R_2*R_{7}*I_3^{-1}}{R_2+R_7}x_4$
A: $$e_2 = C_4^{-1}x_1+C_10^{-1}x_2-R_7I_3^{-1}x_4\color{red}{-R_7R_2^{-1}e_2} \\ e_2\color{red}{+R_7R_2^{-1}e_2} = C_4^{-1}x_1+C_{10}^{-1}x_2-R_7I_3^{-1} \\
\color{blue}{(1+R_7R_2^{-1})}e_2 = C_4^{-1}x_1+C_{10}^{-1}x_2-R_7I_3^{-1}x_4 \\
e_2 = \frac{C_4^{-1}}{\color{blue}{(1+R_7R_2^{-1})}}\color{red}{\frac{R_2}{R_2}}x_1 + \frac{C_{10}^{-1}}{\color{blue}{(1+R_7R_2^{-1})}}\color{red}{\frac{R_2}{R_2}}x_2-\frac{R_7I_3^{-1}}{\color{blue}{(1+R_7R_2^{-1})}}\color{red}{\frac{R_2}{R_2}}x_4 \\
e_2=\frac{C_4^{-1}R_2}{R_2+R_7}x_1+\frac{C_{10}^{-1}R_2}{R_2+R_7}x_2-\frac{R_3I_3^{-1}R_2}{R_2+R_7}x_4$$
