math question : 1/A + 1/B +1/C = 3/7?

Question

1/A + 1/B +1/C = 3/7
A, B, C are all different numbers.
I know the answers. A=4, B=7,C=28, but I don't know the steps.
Is there a formula or a logical way to solve this problem?

Answer 1

there is a logical way to indicate solutions

find smallest A such that 1/ A < 3/7
A= 4
Now if A = 4
1/B+1/C = 3/7 -1/4 = 5/28

5/28 = (1+4)/28 = 1/28+1/7
(2+3)/28 no
if we take A = 5 we may get more solutions

1/B+1/C= 3/7-1/5= 8/35
= (1+7)/35 = 1/35+1/5 allow if duplicates permitted else no
no other

take A =6 and we get

1/B+1/C = 3/7-1/6 = 11/42
no other solution as 11/42 is not sum of reciprocal

Answer 2

the LCD will be 7 or multiple/s of 7
it cannot be 7 since the A,B,C are different
so it can be 6/14 or 9/21 or 12/28
but the denominator has got to be LCD of 4,7 and 28
so it has to be 12/28
so now it has to be 4/28,7/28 and 1/28
so A=4,B=7 and C=28

Answer 3

Normally you 'clear the denominator' by multipling by the product of the denoms (ABC)

-> BC+AC+AB=3ABC/7 , but I don't see how that helps here.l

How about reciprocating the whole thing:

1/[1/A+1/B+1/C] = 7/3

[A+B+C]/ABC = 7/3
[A+B+C]/(ABC/3) = 7 , which says that whatever ABC is it must have
Possible choices are
A,B,C, ABC,
1,3,7, 21 , no sol
1,4,7, 28, -< bingo....

So, the trick is to notice sum must have 7 as a factor, and the product, ABC, must have 3 as a factor.

Answer 4

supply b=Ba, B is a huge decision. c=Ca, C is a huge decision (a million/a)+(a million/b)+(a million/c)=a million/(a+b+c) (a million/a)+(a million/(Ba))+(a million/(Ca))=a million/(a+(Ba)+(Ca)... (a million/a) * (a million+a million/B+a million/C) = (a million/a) * (a million/(a million+B+C)) a million+a million/B+a million/C = a million/(a million+B+C) (BC+C+B)/(BC) = a million/(a million+B+C) (BC+C+B) * (a million+B+C) = BC BC+B²C+BC²+C+BC+C²+B+B²+BC = BC BC+B²C+BC²+C+BC+C²+B+B²+BC-BC = 0 (B²+C²+2BC)+(B²C+BC²)+(B+C) = 0 (B+C)²+BC(B+C)+(B+C)=0 (B+C)( (B+C)+BC+a million )=0 (B+C)(B+a million)(C+a million)=0 So it would want to discover right fee of B,C B = -C or -a million C = -B or -a million aB = -Ca or -a aC = -Ba or -a then b = -c or -a ==> (b+c)=0, (b+a)=0 c = -b or -a ==> (c+b)=0, (c+a)=0 that advise (a+b+c) = a or -a a million/(a+b+c) = a million/a or -a million/a and (a million/a)+(a million/b)+(a million/c) = a million/a or -a million/a then (a million/(a^5))+(a million/(b^5))+(a million/(c^5)) = a million/(a^5) or -a million/(a^5) or (a million/(a^5))+(a million/(b^5))+(a million/(c^5)) = a million/(a+b+c)^5 answer