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I am trying to solve an equation with a constraint involving not-equal.

For example:

$2 x + 3 y = 5$ with the constraint $2 x + 4 ≠ 2$.

How should I approach this?

Edit 1

I tried the suggestions in the comments. I am actually trying to get Mathematica to return a numerical set of values for each variable (which doesnt seem to work via the suggested methods). Any suggestions on this?

Edit 2:

I should have clarified. What I would like Mathematica to do is to give a range of values for both variables $x$ and $y$.

Reduce[{2 x + 3 y == 5, 2 x + 4 != 2}, {y}, Reals]

gives

(x < -1 || x > -1) && y == 1/3 (5 - 2 x)

So there is x-range, but how to get y-range?

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    $\begingroup$ != means not equal. $\endgroup$
    – AsukaMinato
    Oct 29, 2020 at 12:15
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    $\begingroup$ You would write: Solve[{2 x + 3 y == 5, 2 x + 4 != 2}, x] $\endgroup$
    – Daniel Huber
    Oct 29, 2020 at 12:45
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    $\begingroup$ Reduce[{2 x + 3 y == 5, 2 x + 4 != 2}, x] would be better. $\endgroup$
    – Sjoerd Smit
    Oct 29, 2020 at 12:47
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    $\begingroup$ What numerical expression would you find acceptable? $\endgroup$
    – Michael E2
    Oct 29, 2020 at 18:26
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    $\begingroup$ I was wondering if Mathematica could give something like x =/= -1 and y =/= 7/3 or something along the lines of that. $\endgroup$
    – Adrian Lee
    Oct 30, 2020 at 14:02

1 Answer 1

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How about

Reduce[{2 x + 3 y == 5, 2 x + 4 != 2}, {x, y}, Reals]
(*(x < -1 || x > -1) && y == 1/3 (5 - 2 x)*)

?

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  • $\begingroup$ Thanks. I tried that. It looks good - do you think Mathematica could give an explicit set of values that y can take? Or can it only find y in terms of x? Thanks! $\endgroup$
    – Adrian Lee
    Oct 30, 2020 at 14:03

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