System 3

$\large\begin{cases} x^2+y^2+xy+2y+x=2 \,\,\,\,\,\,\,\,\,\,\,...(i) \\ 2x^2-y^2-2y-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{cases}$ Add $(i)$ and $(ii)$ to get $y\,=\,\dfrac{4-x-3x^{2}}{x}$ , or, $\dfrac{(4+3x)(1-x)}{x}\,....(iii)$ .

Put this value of $y$ in $(ii)$ to get, $7x^{4}-23x^{2}+16\,=\,0 \,\implies\, x^{2}\,=\,1\,,\,\dfrac{16}{7}\,\implies\, x\,=\,\pm 1,\,\pm \dfrac{4\sqrt{7}}{7}$ .

Put the obtained value of $x$ in $(iii)$ to get corresponding value of $y$ . Doing so gives following pair of solutions :- $(1,0),\,(-1,-2),\,\left(\dfrac{4\sqrt{7}}{7},\dfrac{-5\sqrt{7}-7}{7}\right),\,\left(\dfrac{-4\sqrt{7}}{7},\dfrac{5\sqrt{7}-7}{7}\right)$ . So, $\left|\displaystyle\sum _{m=1}^{n} x_m+y_m\right|\,=\,\left(1-1+\dfrac{4\sqrt{7}}{7}-\dfrac{4\sqrt{7}}{7}\right)\,+\,\left(0-2-\dfrac{5\sqrt{7}+7}{7}+\dfrac{5\sqrt{7}-7}{7}\right)\,\implies\,|-4|\,\implies\,\boxed{4}$ .