Simultaneous equation 1

From $xy+2x+3y = 10$ , we have $(x+3)(y+2) = 16$ . Let $u=x+3$ and $v=y+2$ , then $uv=16$ , and:

$\begin{aligned} \frac 1{(x+2)(x+4)} + \frac 1{(y+1)(y+3)} & = \frac 2{15} \\ \frac 1{(u-1)(u+1)} + \frac 1{(v-1)(v+1)} & = \frac 2{15} \\ \frac 12 \left(\frac 1{u-1} + \frac 1{v-1} - \frac 1{u+1} - \frac 1{v+1} \right) & = \frac 2{15} \\ \frac 12 \left(\frac {u+v-2}{uv-u-v+1} - \frac {u+v+2}{uv+u+v+1} \right) & = \frac 2{15} & \small \blue{\text{Since }uv=16} \\ \frac 12 \left(\frac {w-2}{17-w} - \frac {w+2}{17+w} \right) & = \frac 2{15} & \small \blue{\text{and let }w=u+v} \\ \frac {w^2-34}{17^2-w^2} & = \frac 2{15} \\ \implies w^2 & = 64 \\ w & = \pm 8 & \small \blue{\text{2 solutions}} \end{aligned}$

Therefore, there are $\boxed 2$ solutions, that is

$\begin{aligned} (u+v)^2 & = 64 \\ u^2 + 2uv + v^2 & = 64 \\ u^2 + \frac {16^2}{u^2} & = 32 \\ u^4 - 32u^2 + 256 & = 0 \\ (u^2 - 16)^2 & = 0 \\ \implies u & = \pm 4 \implies v = \pm 4 \\ \implies (x,y) & = (1, 2), (-7, -6) \end{aligned}$