## Calculus with Applications (10th Edition)

$\dfrac{1}{m-1}+\dfrac{2}{m}=\dfrac{3m-2}{m(m-1)}$ or $\dfrac{1}{m-1}+\dfrac{2}{m}=\dfrac{3m-2}{m^{2}-m}$
$\dfrac{1}{m-1}+\dfrac{2}{m}$ Evaluate the sum of these two fractions by using the LCD, which is $m(m-1)$ in this case: $\dfrac{1}{m-1}+\dfrac{2}{m}=\dfrac{m+2(m-1)}{m(m-1)}=...$ Simplify: $...=\dfrac{m+2m-2}{m(m-1)}=\dfrac{3m-2}{m(m-1)}$