solve for d: a(d-b)=db-c?

Answer 1

Expand it...

So:

a(d-b)=db-c
ad-ab=db-c

Then group the d terms together on the left...

ad-db=ab-c
d(a-b)=ab-c
d = (ab-c)/(a-b)

And that's the sol'n (without knowing a, b, or c)
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Answer 2

Actually none of the previous answers is right

I agree until here

a(d-b) = db- c
ad - bd = ab - c

d(a-b) = ab-c

*BUT* you can only find d if a is different from b, otherwise you would have 0d = a^2 - c

And then 0 = a^2 - c which wouldn't make any sense if a^2 would be equals to c (and you would get equalities like 0d = 5 or so, which leads you to 0 = 5) and which would give you only 0d = 0 if a^2 = c. And any d you would plug hier would give you 0 = 0.

Ana

Answer 3

ad-ab = db-c
ad-db = ab-c
d(a-b) = ab-c
d = (ab-c)/(a-b)

Answer 4

a(d-b)=db-c
ad-ab=db-c
ad-db=ab-c
d(a-b)=ab-c
d=(ab-c)/(a-b)

Doug