How do you solve for d in #n= (dh)/(f+d)#?

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Answer 1 · 2016-07-17T01:25:22

#color(green)((nf)/(h-n) = d)#

To solve for #d#, we need to get it by itself (isolate the variable). First, undo the division by multiplying both sides by the denominator.

#n = (dh)/(f+d)#

#(f+d)/1 * n/1 = (dh)/cancel(f+d) * cancel(f+d)/1#

#((f+d)timesn) = dh#

#nf + nd = dh#

Now get all terms with #d# to one side and factor out the #d#.

#nf = dh-nd#

#nf = d(h-n)#

Finish isolating the variable by getting #d# by itself.

#(nf)/(h-n) = (d(cancel(h-n)))/cancel(h-n)#

#color(green)((nf)/(h-n) = d)#

This is your final answer.