How do you solve #T>(L+D)/B# for #D#?

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Answer 1 · 2016-07-29T14:00:40

#D<TB-L#

In order to solve for #D# in the equation #T>(L+D)/B#

Begin with the multiplicative inverse and multiply by #B# on both sides.

#T(B)>(L+D)/cancelB (cancelB)#

#TB> L+D#

Now use the additive inverse to isolate the #D#

#TB - L > cancelL+D cancel(-L)#

#TB-L>D#