2
$\begingroup$

How can one write $\frac{ad+bc}{bd}$ in $\frac{a}{b}+\frac{c}{d}$ form? In other words, is there something opposite of Together[]?

Edit

My intend was to write

(8380*(x6[t] - x7[t])*(-0.130610 + Sqrt[(x6[t] - x7[t])^2 + (y6[t] - y7[t])^2]))/
   Sqrt[(x6[t] - x7[t])^2 + (y6[t] - y7[t])^2]

as

8380*(x6[t] - x7[t])*(1 - 0.13061/Sqrt[(x6[t] - x7[t])^2 + (y6[t] - y7[t])^2])
$\endgroup$
  • $\begingroup$ FullSimplify[(ad + bc)/(b*d)] $\endgroup$ – Fraccalo Aug 18 '19 at 13:37
  • 1
    $\begingroup$ Also look at Apart as the counterpart of Together. For example, Apart[(a*d + b* c)/(b*d)]. $\endgroup$ – Tim Laska Aug 18 '19 at 14:08
  • $\begingroup$ I used both these functions for the expression given in Edit. Neither of them worked. $\endgroup$ – Soumyajit Roy Aug 18 '19 at 14:56
  • 1
    $\begingroup$ 8380*(x6[t]-x7[t])*FullSimplify[(-0.130610 + Sqrt[(x6[t]-x7[t])^2 + (y6[t]-y7[t])^2]) /Sqrt[(x6[t]-x7[t])^2 + (y6[t]-y7[t])^2]] instantly gives me 8380*(x6[t]-x7[t])*(1-0.13061/Sqrt[(x6[t]-x7[t])^2 + (y6[t]-y7[t])^2]) $\endgroup$ – Bill Aug 18 '19 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.