0
$\begingroup$

I have outputs like "expression/(1+3*c^2) +anotherexpression/(1+3*c^2)" and so on.

Is there any way to set 1/(1+3*c^2)=A where A is a new variable so that the output reads like "A*expression+A*anotherexpression"?

Thanks

$\endgroup$
1
  • $\begingroup$ TransformationFunctions or you can just use (n + m + k)/(1 + 3*c^2) + x/(1 + 3*c^2) /. {1/(1 + 3 c^2) -> A} $\endgroup$
    – Sektor
    Jul 31 '14 at 10:02
0
$\begingroup$
ReplaceAll[x/(1 + 3 c^2) + y/(1 + 3 c^2), 1/(1 + 3 c^2) -> z]

x z + y z

$\endgroup$
1
  • $\begingroup$ Works fine. Thank you very much :) $\endgroup$
    – Trac3
    Jul 31 '14 at 10:10

Not the answer you're looking for? Browse other questions tagged or ask your own question.