My goal is to add three equations: a==b, c==d, e==f. I could used MapThread and Replace as shown below, but is there a simpler method?
eqs={a==b, c==d, e==f};
MapThread[Plus,eqs/.Equal->List]/.List->Equal (* a+c+e==b+d+f *)
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4 Answers
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1
eqs = {a == b, c == d, e == f};
<< EqualThread.m
Total[eqs]
(*a+c+e==b+d+f*)
EqualThread is found at https://library.wolfram.com/infocenter/ID/4491/, and it has worked for every version of Mathematica since the 90's. If you put it in your init.m file, adding, subtracting, multiplying and dividing equations becomes effortless.
It also works for functions, such as
Log[eqs[[1]]]
(*Log[a] == Log[b]*)
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$\begingroup$ Thank you for your answer; I have learned that Equal funtion have not Listable property for some reason. And as the use of TagSetDelayed, Unprotected and especilly, the Thread function in the little packge can handle this case also. $\endgroup$Soon– Soon2025-06-19 07:27:31 +00:00Commented Jun 19 at 7:27
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Option 1
eqs = {a == b, c == d, e == f};
Equal @@ Total[List @@@ eqs]
Option 2
Equal @@ Last@Accumulate[List @@@ eqs]
Option 3
Thread[Total[eqs],Equal]
Unfortunately, AddSides works on only pair of equations and not set of equations. Hence this works
AddSides[a == b, c == d]
But not
AddSides[a == b, c == d, e == f]
But you can do
option 4
Last@FoldList[AddSides,eqs]
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9$\begingroup$ for option 4, just
Fold[AddSides, eqs]suffices. $\endgroup$lericr– lericr2025-06-17 15:28:10 +00:00Commented Jun 17 at 15:28 -
1$\begingroup$ @lericr I really was thinking of trying
Foldbut for some reason, I forgot. Good it works. $\endgroup$Nasser– Nasser2025-06-17 15:50:59 +00:00Commented Jun 17 at 15:50 -
$\begingroup$ Thank you for your reply. I've learned that the functions for repetitive or nested computation can also be used cleverly in this context. $\endgroup$Soon– Soon2025-06-18 12:29:01 +00:00Commented Jun 18 at 12:29
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Try this:
eqs = {a == b, c == d, e == f};
Equal @@ Plus @@@ Transpose[List @@@ eqs]
(* a + c + e == b + d + f *)
Have fun!
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1
eqs = {a == b, c == d, e == f};
Total[eqs[[All, 1]]] == Total[eqs[[All, 2]]]
(* a + c + e == b + d + f *)
```
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1$\begingroup$ (+1) Also
ApplySides[Total, #[[All, 1]] == #[[All, 2]]] &[eqs]? $\endgroup$user1066– user10662025-06-19 14:50:33 +00:00Commented Jun 19 at 14:50