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When the unknown is a name which is only used as a function in the equation, the solver constructs a function or procedure which solves the equation.

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

$x\mapsto x-2$

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

$x-2$

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

$x\mapsto x-2$

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

$x-2$

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

$x\mapsto x-2$

$\mathrm{solve}\left({f\left(x\right)}^4-{f\left(x\right)}^3+x+1\,f\right)$

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