$I+J=R$ and $r+s=1, r\in I,s\in J$ then $sx+ry\in IJ\Rightarrow x\in I$ and $y\in J$

Question

Let $R$ be a commutative ring with unity. $I+J=R$ with $I,J$ Ideals and $r+s=1, r\in I,s\in J$ then $sx+ry\in IJ\Rightarrow x\in I$ and $y\in J$.

It should be very obvious. How can I conclude that $sx\in I$, if $sx\in I$ then $x\in I$ by $s=1-r$. If I reduce by $IJ$ I get $sx+ry=s(i_x+j_x)+r(i_y+j_y)=(1-r)j_x+(1-s)i_y=j_x+i_y$ with $i_x+j_x=x$ and $i_y+j_y=y$ and $i_p\in I, j_p\in J$. I can't see why I can say $j_x+i_y=0$.

Answer 1

$sx+ry= sx+rx-rx+ry= x(s+r)+r(y-x) =x+r(y−x) \in IJ\subset I$
on the other hand $r(y−x) \in I$
so $x\in I$
similarly $y\in J$