# Math Help - a question......

1. ## a question......

$a.b-c=4$

$b.c-a=3$

in equality, a, b, c are positive integers.

What is total of $a+b+c$ minumum?

2. $a.b-c=4$

$b.c-a=3$

$ab+bc-a-c=7$

move things around:

$ab-a+bc-c=7$

factor:

$a(b-1)+c(b-1)=7$

factor again:

$(a+c)(b-1)=7$

hmm, 7 is a prime number though and so cannot have two integer factors, unless $b-1 = 1$ and $a+c=7$ or $b-1=7$ and $a+c=1$ (which means that in this case either a or c must be zero, I don't know if you consider that a positive integer)

In the first case: $b = 2$ and $a+c=7$

So if we add those two together: $a+b+c=9$

In the second case: $b = 8$ and $a+c=1$

So if we add those two together: $a+b+c=9$

So the answer is: $\boxed{a+b+c=9}$