Answer
$\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}=\dfrac{23}{20(k-2)}$
or
$\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}=\dfrac{23}{20k-40}$
Work Step by Step
$\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}$
Evaluate the sum of these two fractions by using the LCD, which is $(5)(4)(k-2)$ in this case:
$\dfrac{2}{5(k-2)}+\dfrac{3}{4(k-2)}=\dfrac{(2)(4)+(3)(5)}{(5)(4)(k-2)}=...$
Simplify:
$...=\dfrac{8+15}{20(k-2)}=\dfrac{23}{20(k-2)}$
