How do i work out what B is from the following formula
A = B / B - C
many thanks
2007-04-05 20:59:29 · 9 answers · asked by idai 5 in Science & Mathematics ➔ Mathematics
i assume you mean A= B/(B-C).
cross multiplication: AB-AC = B
subtract AB from both sides: -AC = B-AB
multiply each side by -1: AC = AB-B
take out common factor from RHS: AC = B(A-1)
divide each side by (A-1); therefore B = AC/(A-1).
of course, i'm assuming you know the values of A and C; otherwise there would really be no way to solve 3 unknowns using only one equation, since you need to use simultaneous equations (or sth of the sort).
2007-04-05 21:11:38 · answer #1 · answered by rc 2 · 0⤊ 0⤋
Rewrite
A = B/{B-C} [Eq 1]
To isolate a term in an equation you MUST REMEMBER what you do to One side you MUST ALSO do to the Other Side
To 'remove' [B-C] from RHS multiply both Dides by [B-C]
Rewrite A{B-C} = B [Eq 2]
Factorise LHS AB - AC = B
Collect terms with B and Rearrange
AB - B = AC [Eq 3]
Simplify LHS and rewrite
B{A - 1} = AC [Eq 4]
'Remove' {A-1} from LHS
B = AC/[A-1]
The limits of the formatting in this portal do not allow any clearer explanation - I hope this hasworked for you :-)
2007-04-06 08:41:59 · answer #2 · answered by Rod Mac 5 · 1⤊ 0⤋
No way to tell. The only thing we can be sure of is that B does not equal 0. (Division by zero is undefined).
A = B/B - C = 1 - C.
As you can see, the Bs fall out.
The others all assumed you wrote your question wrong.
A + C = B/B
A + C = 1
B could be anything except zero.
2007-04-06 04:18:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
U can find value of B as under
A=B/B-C
Multipy the equation by "B" on both sides we get
AB=B-BC We can write this equation in the following manner
AB-B+BC=0 Here B is common it can be safely taken out
B(A-1+C)=0
B= 0/(A-1+C)
B=0
2007-04-06 04:42:53 · answer #4 · answered by diamond 3 · 0⤊ 0⤋
If you mean the parenthesis formula they've already shown you, well, they've shown you the solution as well.
If you really mean: A = B / B - C, then, B could be any number DIFFERENT TO 0 (as you can't divide by 0), because any number divided by itself is 1.
2007-04-06 05:07:09 · answer #5 · answered by pablo_cg86 3 · 0⤊ 0⤋
A= B/B-C
Therefore 1-A = 1 - B/(B-C) = (B-C -B)/(B-C)= -C/(B-C)
therefore by cross multiplication & interchanging
(B-C)= -(C)/(1-A)
= C/(A-1)
Or B = C + C/(A-1) = C{ 1 + 1 /A-!} = C.A/(A-1) = A.C/(A-1)
or A.C/(A-1)
2007-04-06 04:15:21 · answer #6 · answered by RAJASEKHAR P 4 · 1⤊ 0⤋
You have to get B alone
A+C=B/B
anything divided by its self becomes one(not zero)
A+C=1
2007-04-06 04:36:06 · answer #7 · answered by Anonymous · 0⤊ 0⤋
I'm assuming you mean A = B / (B - C)
multiply both sides by (B - C)
(B - C) A = B
AB - AC = B
AB - B = AC
B (A - 1) = AC
B = AC / (A-1)
2007-04-06 04:11:49 · answer #8 · answered by Demiurge42 7 · 1⤊ 0⤋
I assume you mean A = B / (B-C)...
So multuply by (B-C)... AB - AC = B
Get all Bs on left... (A-1)B = AC
Dividing... B = AC / (A - 1)
2007-04-06 04:09:14 · answer #9 · answered by Anonymous · 1⤊ 0⤋