5
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Say I have the following input:

Reduce[{t*c==50, t==5}]

the output:

t == 5. && c == 10.

Is there a way to create a variable x that gets assigned the value of c from the output without actually having to type x = 10?

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  • 6
    $\begingroup$ x = c /. ToRules @ Reduce[...] $\endgroup$
    – wxffles
    Sep 9, 2014 at 21:00

3 Answers 3

2
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out = Reduce[{t*c == 50, t == 5}];
x = Cases[out, c == y_ :> y, -1][[1]]
10
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With

eqs = {t*c == 50, t == 5};

then

x = Refine[c, Reduce[eqs]]

or (wxffles' comment)

x = c /. ToRules@Reduce[eqs]

or

x = c /. First@Solve[eqs]
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From the documentation to Solve (just ran into it):

Solve with Method->"Reduce" uses Reduce to find solutions, but returns replacement rules:

Solve[{t*c == 50, t == 5}, {t, c}, Method -> Reduce]

{{t -> 5, c -> 10}}
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