solve for sum of variables? Given a=bc+de, can I solve for b+d=(something w/o b or d)?

[Adamos]

Good day. I have equation like this a=bc+de and I wonder whether i can solve it for b+d = something not including b and d. Is it possible? If it is how can i do it? Thank you in advance.

 

[topsquark]

In a word, Nope.

-Dan

 

[Adamos]

May i ask why it is not possible?

 

[topsquark]

There is no way to isolate both b and d. No matter what you do b will be a function of d (or d is a function of b.) The best you could do would be something like this:
\(\displaystyle a = bc + de\)

\(\displaystyle \dfrac{a}{c} = b + \dfrac{de}{c}\)

\(\displaystyle b = \dfrac{a}{c} - \dfrac{de}{c} = \dfrac{1}{c} \left ( a - de \right )\)

Then add d to both sides:
\(\displaystyle b + d = \dfrac{1}{c} \left ( a - de \right ) + d = \dfrac{1}{c} \left ( a - de + dc \right )\)

No matter what you do you can't get rid of the d on the RHS.

-Dan

 

[Otis]

Of course, whenever c=e then:

b+d = a/c

?

 

[Steven G]

There is no way to isolate both b and d. No matter what you do b will be a function of d (or d is a function of b.) The best you could do would be something like this:
\(\displaystyle a = bc + de\)

\(\displaystyle \dfrac{a}{c} = b + \dfrac{de}{c}\)

\(\displaystyle b = \dfrac{a}{c} - \dfrac{de}{c} = \dfrac{1}{c} \left ( a - de \right )\)

Then add d to both sides:
\(\displaystyle b + d = \dfrac{1}{c} \left ( a - de \right ) + d = \dfrac{1}{c} \left ( a - de + dc \right )\)

No matter what you do you can't get rid of the d on the RHS.

-Dan

Are you 100% about what you said??

 

[topsquark]

Except for the case c = e yes, I am 100% sure.

(Don't embarass me Mommy! ? )

-Dan