Answer
$\dfrac{5}{2r+3}-\dfrac{2}{r}=\dfrac{r-6}{r(2r+3)}$
or
$\dfrac{5}{2r+3}-\dfrac{2}{r}=\dfrac{r-6}{2r^{2}+3r}$
Work Step by Step
$\dfrac{5}{2r+3}-\dfrac{2}{r}$
Evaluate the subtraction of these two fractions by using the LCD, which is $r(2r+3)$ in this case:
$\dfrac{5}{2r+3}-\dfrac{2}{r}=\dfrac{5r-2(2r+3)}{r(2r+3)}=...$
Simplify:
$...=\dfrac{5r-4r-6}{r(2r+3)}=\dfrac{r-6}{r(2r+3)}$
