Fraction No More!

Since the denominator is even, this means that the numerator must be even, which tells us that y is odd. $\frac{5y^{2}-9}{2y+6}$ = $\frac{5y^{2}-45+36}{2y+6}$ = $\frac{5(y^{2}-9)+36}{2(y+3)}$ = $\frac{5(y+3)(y-3)+36}{2(y+3)}$ = $\frac{5(y+3)(y-3)}{2(y+3)}$ + $\frac{36}{2(y+3)}$ = $\frac{5(y-3)}{2}$ + $\frac{18}{y+3}$ Because y is odd $\frac{5(y-3)}{2}$ and $\frac{18}{y+3}$ are integers (y+3)|18 so y could be:±1,±2,±3,±6,±9 and ±18
We know that y is odd so y+3 is even so
y+3=2
y+3=-2
y+3=6
y+3=-6
y+3=18
y+3=-18
Finally we get six values of y:{-21,-9,-5,-1,3,15}. So there are 6 such ys.

Clearly, $y$ must be odd, $y=2x+1$ , so that $\frac{5y^2-9}{2y+6}=5x-5+\frac{9}{x+2}$ . Thus $x+2=\pm 1, \pm 3$ or $\pm 9$ . There are $\boxed{6}$ solutions for $x$ and $y$ .

I got first answer from TI -83 PLUS and then confirmed by method below.

$\dfrac{5y^{2}-9}{2y+6}= \frac 1 2 \left (\dfrac{(5y^{2}+15y) - (15y+45) +36}{y+3} \right ) = \frac 1 2 \left (5y - 15) \right )+\dfrac {18}{y+3}.\\ If\ y\ is\ even,\\ \therefore \ \dfrac{5y- 15} 2=\dfrac{odd} 2,\ so\ \dfrac {18}{y+3} \ must\ also\ be\ =\dfrac{odd} 2.\\ \implies\ \dfrac {18}{y+3}= \ \frac 3 2\ \ or\ \ \ \frac 9 2.\\ \implies\ 36=3(y+3) \ \ OR\ \ 36=9(y+3). \ \ Both \ not\ possible\ with \ even\ y.\ So\ y\ is\ not\ even.\\ If\ y\ is\ odd,\\ \therefore \ \dfrac{5y-15} 2 \text{ is an integer. ............ So (y+3) |18 with odd y.} \\ \implies\ \text{( y+3) is even. And 18=2*odd. So only odd factors of 18 to be considered.}\\ \text{Odd Factors of 9:- \{-9, -3. -1, 1, 3, 9\}}\\ \text{Corresponding y:- {-5, -9, -21, 15, 3, -1} }\\ {\Large \color{#D61F06}{6} } \ \ values\ of\ y.$

After Long Division,

$\frac{5y^2-9}{2y+6}=\frac{5}{2}(y-3)+\frac{18}{y+3}$

$2y+6$ is even $\Rightarrow 5y^2-9$ is even $\Rightarrow 5y^2$ is odd $\Rightarrow y$ has to be odd.

Since $y$ is odd, $\frac{5}{2}(y-3)$ is an integer $\therefore \frac{18}{y+3}$ is an integer as well. $\therefore y+3$ is an even integer factor of 18, since $y$ is odd as stated above.

$\because 18=1\times 18=2\times 9=3\times6$

We have $\pm 2,\pm 6,\pm 18$

$\therefore\boxed{6}$ integer solutions of $y$ correspondingly.

Using partial fractional decomposition, the expression becomes; $\frac { 5(y-3) }{ 2 } +\frac { 36 }{ 2y+6 }$

Thus, $y$ has to be odd and the rest could be done by simple substitution of integers.