1/A=1/B+1/C....?Help me with this answer please where you solve for A?

Question

I need to solve for A

would the answer be A=3+B+C? Or what? HELP!! THanks

Answer 1

First, multiply everything by the LCD, in this case, it's ABC:

BC = AC + AB

Now factor out the A:

BC = A(C + B)

Now divide by (C+B)

A = BC/(C + B)

Answer 2

First, you add 2 fractions 1/A and 1/B (I'm sure you know how to do it).

The result you get is equal to 1/A, as given in the problem.

Now to have A, which is the inverse of 1/A, take the inverse of the result you got above. Done!

Answer 3

Do the same thing to both sides of the equasion

multiply by a   1= a(1/b+1/c)

divide by (1/b+1/c)     a= 1/(1/b+1/c)

Answer 4

there must be some other piece of information plz put all details

Answer 5

1/A = 1/B + 1/C

=> 1/A = (B+C)/BC

=> A = BC / (B+C)

Answer 6

multiply   the whole eqtn with  ABC

ABC( 1/A= 1/B+ 1/C)

BC= AC+ AB

BC= A( C+B)

A( C+B)= BC

A= BC/ ( C+B)