Treat 2 variable as 1

Question

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]

What about /. x -> z? Notice that x z is a multiplication. — Kuba, Jul 07 '17 at 08:28
@Kuba not sure, what you're getting at. Perhaps one could streamline OP's workaround somewhat like Solve[equations /. x -> xz/z, xz] /. xz -> x z taking out the not always reliable Expand — LLlAMnYP, Jul 07 '17 at 08:31
@LLlAMnYP I understood that OP does not want to define a relation but to consider x and z the same variable, therefore I suggested that. Then one can solve for z as there is only z for z or x. — Kuba, Jul 07 '17 at 08:35
@Kuba I should have been more clear. I was asking if there was a way to treat xz as a new variable, one that is different from both x or z. — I should change my Username, Jul 07 '17 at 08:38
@LLIAMnYP Is there anyway that this could be done without using replacements? — I should change my Username, Jul 07 '17 at 08:38
OTOH that wouldn't work, a second equation seems to require a second variable to solve for, i.e. Solve[..., {x, var}] or such. — LLlAMnYP, Jul 07 '17 at 08:44
@Kuba or, in other words, for a b c - d e f == 0 we have no idea what a, b, and c are, but we know that their product, a b c is equal to d e f. — LLlAMnYP, Jul 07 '17 at 08:47
@Kuba it's a nice question, but quite sure, it's a dupe, maybe not of your link, but of those therein, what do you think? — LLlAMnYP, Jul 07 '17 at 08:58
@LLlAMnYP maybe one of those links is better but the question asks about replacing z x with var while the linked one a / b with k, I'd say it is close enough. But I don't have time to focus on details so I will leave the link for people judgement. — Kuba, Jul 07 '17 at 09:03
why do you need to do it without using replacements? Because that seems the natural method to do this kind of things — glS, Jul 07 '17 at 11:07
@glS I was just wondering if there was another or a shorter way to do it. I'm familiar with replacements so I am curious if the same behavior could be achieve with something I didn't know about — I should change my Username, Jul 07 '17 at 11:21
well at least in this case Solve[eq == 0, HoldPattern@Times[x, z]] works, but I don't know how stable of a solution this would be in more complex cases — glS, Jul 07 '17 at 13:01

score 2 · Answer 1 · answered Jul 07 '17 at 15:18

2

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}]

answered Jul 07 '17 at 15:18

Sjoerd Smit

23,370
46
75

score 0 · Answer 2 · answered Jul 07 '17 at 14:39

0

Perhaps:

eq = Expand[x (z+y)];
Reduce[eq == 0 && v == x z, v]

(x == 0 && v == 0) || (y == -z && v == x z)

answered Jul 07 '17 at 14:39

Carl Woll

130,679
6
243
355

Treat 2 variable as 1

2 Answers2