Let $R$ be a ring with unity and $I$ be a proper ideal of $R$. If for some $a,b\in R$,
$ab=0_R=ba$ and
$yb\in I$ for some $0\neq y\in R$.
By stating suitable conditions on $R$ or $I$, prove that $y-ra\in I$ for all $r\in R$.
This is my attempt on the proof:
Since $ba=0=ab$, then $(y-ra)b=yb-(ra)b=yb-r(ab)=yb-r0=yb\in I$. So that $y-ra\in (I:_Rb)=\{s\in R:sb\in I\}$. But this does not help me much.