## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\frac{IR}{E-Ir}=n$
$I=\frac{nE}{R+nr}$ ; n We are trying to solve for the variable n. We must first multiply both sides by $(R+nr)$. We are doing this to cancel out the fraction. $I\times(R+nr)=\frac{nE}{R+nr}\times(R+nr)$ $IR+Inr=nE$ We now subtract Inr from both sides. $IR+Inr-(Inr)=nE-(Inr)$ $IR=nE-Inr$ From here we factor out n. $IR=n(E-Ir)$ Lastly, you divide both sides by $(E-Ir)$. $\frac{IR}{E-Ir}=n\frac{E-Ir}{E-Ir}$ The fraction will cancel itself out, giving us our answer. $\frac{IR}{E-Ir}=n$