Question 1097131
.
Below find the &nbsp;<U>CORRECT</U>&nbsp; solution.


<pre>
{{{E(R-r)}}} = {{{e(R+r)}}}

{{{ER-Er}}} = {{{eR+er}}}

{{{-er-Er}}} = {{{-ER+eR}}}

{{{-r(e+E)}}} = {{{R(-E+e)}}}

{{{r(e+E)}}} = {{{R(E-e)}}}

r = {{{R((E-e)/(E+e))}}}
</pre>