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Can I use the output of reduce to define a variable in a function definition? In other words, can I somehow skip the step where I manually enter t:=2?

Reduce[{-2 + t == 0,3 t == 6,4 + t == 6},t] t==2 t := 2 f[x_, y_, z_] = {-2 + t, 3 t, 4 + t} {0, 6, 6}

I tried ./ and ToRule, neither seem to work

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  • $\begingroup$ ClearAll[t]; sol = ToRules@Reduce[{-2 + t == 0, 3 t == 6, 4 + t == 6}, t]; $\endgroup$ Commented Mar 26, 2018 at 22:47
  • $\begingroup$ that's... interesting. I wish there was some less convoluted technique, but if y'all decide which one of you wants to write an answer I will accept it. $\endgroup$ Commented Mar 26, 2018 at 23:13

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ClearAll[t]; t=t/.ToRules[Reduce[{-2 + t == 0,3 t == 6,4 + t == 6},t]]
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