Is there anyway to treat two variable, that are a result of expansion, as one?
To use Solve[]
I can do
Clear[x, z, y, eq]
eq = Expand[x (z + y)] /. x z -> xz
Solve[eq == 0, xz]
Is there anyway to solve directly like
Clear[x, z, y, eq]
eq = Expand[x (z + y)]
Solve[eq == 0, x z]
/. x -> z
? Notice thatx z
is a multiplication. $\endgroup$ – Kuba♦ Jul 7 '17 at 8:28Solve[equations /. x -> xz/z, xz] /. xz -> x z
taking out the not always reliableExpand
$\endgroup$ – LLlAMnYP Jul 7 '17 at 8:31x
andz
the same variable, therefore I suggested that. Then one can solve forz
as there is onlyz
forz
orx
. $\endgroup$ – Kuba♦ Jul 7 '17 at 8:35Solve[eq == 0, HoldPattern@Times[x, z]]
works, but I don't know how stable of a solution this would be in more complex cases $\endgroup$ – glS Jul 7 '17 at 13:01