Answer
$\dfrac{3k}{2k^{2}+3k-2}-\dfrac{2k}{2k^{2}-7k+3}=\dfrac{k(k-13)}{(2k-1)(k+2)(k-3)}$
Work Step by Step
$\dfrac{3k}{2k^{2}+3k-2}-\dfrac{2k}{2k^{2}-7k+3}$
Factor the denominators of both rational expressions:
$\dfrac{3k}{2k^{2}+3k-2}-\dfrac{2k}{2k^{2}-7k+3}=...$
$...=\dfrac{3k}{(k+2)(2k-1)}-\dfrac{2k}{(2k-1)(k-3)}=...$
Evaluate the subtraction of these two fractions by using the LCD, which is $(2k-1)(k+2)(k-3)$ in this case:
$...=\dfrac{3k(k-3)-2k(k+2)}{(2k-1)(k+2)(k-3)}=...$
Simplify:
$...=\dfrac{3k^{2}-9k-2k^{2}-4k}{(2k-1)(k+2)(k-3)}=\dfrac{k^{2}-13k}{(2k-1)(k+2)(k-3)}=...$
$...=\dfrac{k(k-13)}{(2k-1)(k+2)(k-3)}$
