resolve this

a*b=15
b*c=35
a*c=21
a=?
b=?
c=?

first get b in terms of a
now put the new value of b into the second equation and rearraange to get c in terms of a
put new value of c into third equation and solve for a
that should be enough to get you started.

ab = 15 therefore a and b can either be 1 and 15 or 3 and 5
bc = 35 therefore b and c can either be 1 and 35 or 5 and 7
ac = 21 therefore a and c can either be 1 and 21 or 3 and 7
in order for the system of equation to be true, the same number has to appear in multiple equations
therefore 15, 35 and 21 are out making their pair (1) out as well.
Now, ab has a product that is not a multiple of 7 therefore a or b cannot be 7.
bc has a product that is not a multiple of 3 therefore b or c cannot be 3.
ac has a product that is not a multiple of 5 therefore a or c cannot be 5.
So a doesn't equal 5 or 7
b doesn't equal 3 or 7
c doesn't equal 3 or 5
Finally!
a = 3
b = 5
c = 7

Hi reasonforeverything
What about if a,b or c was not a whole number?

a*b=15 <=> a=15/b
b*c=35 <=> c=35/b
a*c=21 <=> 15/b * 35/b = 21 <=> 525/b^2 = 21 <=> b^2 = 525/21 <=> b = sqroot(25) <=> b = 5

Now that you got the value of be it's easy:
a = 15/b <=> a = 15/5 <=> a = 3
c = 35/b <=> c = 35/5 <=> c = 7

There you go: a=3; b=5; c=7