Question 1084727
I'm going to substitute variables to make it more readable and then substitute back at the end.
{{{A=R[x]}}}
{{{B=R[o]}}}
{{{C=R[f]}}}
{{{D=R[i]}}}
{{{E=R[c]}}}
So,
{{{A+B=((C-D)/C)(E+B)}}}
{{{A=((C-D)/C)(C+B)-B}}}
{{{A=((C-D)/C)(C+B)-B(C/C)}}}
{{{A=((C-D)(C+B)-BC)/C)}}}
{{{A=(C^2+BC-CD-BD-BC)/C)}}}
{{{A=(C^2-CD-BD)/C)}}}
{{{A=(C^2-D(B+C))/C}}}
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What are you looking to get to?