$The number of pairs (x, y) satisfying the equations$  $\sin x+\sin y=\sin (x+y)\text{ and }\left|x\right|+\mid$$y\mid =1\text{ is }$

$\begin{array}{l}\text{The first equation can be written as }2\sin \frac{1}{2}(x+y)\cos \frac{1}{2}(x-y)=2\sin \frac{1}{2}\left(x+y\right)\cos \frac{1}{2}\left(x+y\right)\\ \\ \therefore \text{ Either }\sin \frac{1}{2}(x+y)=0\text{ or }\sin \frac{1}{2}x=0\text{ or }\sin \frac{1}{2}y=0\\ \because x+y=1,x+y=-1,x-y=-1,x-y=1\\ \text{ When }x+y=0,\text{ we have to reject }x+y=1\\ x+y=-1\text{ and solve it with }x-y=1\\ \text{ or }x-y=-1\text{ which gives }\left(\frac{1}{2},-\frac{1}{2}\right)\text{ or }\left(-\frac{1}{2},\frac{1}{2}\right)\text{ as the possible solution. }\\ \text{ Again solving with }x=0\text{ , we get }(0,\pm 1)\text{ and solving with }y=0,\text{ we get }\\ (\pm 1,0)\text{ as the other solution. Thus we have six pairs of solutions for }x\text{\hspace{0.17em}and }y\end{array}$