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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]
How about introducing the extra variable as a new equation and then telling Solve to solve for 2 variables? Or use Reduce instead?
Clear[x, z, y, eq];
eq = Expand[x (z + y)];
Solve[{eq == 0, xz == x z}, {xz, y}]
Reduce[{eq == 0, xz == x z}, {xz}]
Perhaps:
eq = Expand[x (z+y)];
Reduce[eq == 0 && v == x z, v]
(x == 0 && v == 0) || (y == -z && v == x z)
/. x -> z? Notice thatx zis a multiplication. $\endgroup$ – Kuba♦ Jul 7 '17 at 8:28Solve[equations /. x -> xz/z, xz] /. xz -> x ztaking out the not always reliableExpand$\endgroup$ – LLlAMnYP Jul 7 '17 at 8:31xandzthe same variable, therefore I suggested that. Then one can solve forzas there is onlyzforzorx. $\endgroup$ – 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