Localisation (Algebra) Well defined

Question

I want to show that the multiplication is well defined in the following operation. We have $R$ a commutative Ring. $(r,s)~(r',s')$ is defined as $\exists \bar{s}:\ \bar{s}(rs'-r's)=0 \ \ (*)$. And the multiplication is defined as $\frac{r}{s}\cdot\frac{r'}{s'}=\frac{rr'}{ss'}$, where $\frac{r}{s}$ is means $(r,s)$.
I know I have to take two "sets" of pairs, so i would have $r,s,r',s',t,u,t',u'$ and then there exists $\bar{s}\bar{s'}$ so that $(*)$. I only have to show now that: $\bar{s}\bar{s'}(rr'uu'-tt'ss')=0$ and I am missing the "trick" or step to get to the $0$.

Answer 1

$\bar{s}\bar{s'}rr'uu'-\bar{s}\bar{s'}tt'ss'-\bar{s}\bar{s'}r'u'ts+\bar{s}\bar{s'}r'u'ts=(ru-ts)\bar{s}(r'u'\bar{s'})+(r'u'-t's')\bar{s'}(ts\bar{s})=0+0=0$