>

$\mathrm{solve}\left(f\left(x\right)x+2\,f\right)$

${x}{\mapsto}{x}{}{2}$
 (1) 
This is more or less equivalent to:
>

$\mathrm{solve}\left(f\left(x\right)x+2\,f\left(x\right)\right)$

>

$\mathrm{unapply}\left(\,x\right)$

${x}{\mapsto}{x}{}{2}$
 (3) 
This in turn is more or less equivalent to:
>

$\mathrm{solve}\left(yx+2\,y\right)$

>

$\mathrm{unapply}\left(\,x\right)$

${x}{\mapsto}{x}{}{2}$
 (5) 
More complicated inputs will generally return answers involving RootOfs
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$\mathrm{solve}\left({f\left(x\right)}^{4}{f\left(x\right)}^{3}+x+1\,f\right)$

${x}{\mapsto}{\mathrm{RootOf}}{}\left({{\mathrm{\_Z}}}^{{4}}{}{{\mathrm{\_Z}}}^{{3}}{+}{x}{+}{1}\right)$
 (6) 