# The number of ordered triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations $ab + bc = 44$, $ac + bc = 33$

My try:

Subtracting the eqns:

$$a(b-c) = 11$$

$$a=1,b-c=11$$ OR $$a=11,b-c=1$$

Substituting these values back int the original eqn. does not give an integral answer.

Thus number of ordered pairs = $$0$$.

Please also tell a general way of approaching problems in which we have to find the number of integral ordered pairs given some equations.

• Why don't you like this solution ? Feb 5 '14 at 17:37
• @Peter I just said I am not confident about it[Not that I don't like it].Is this then correct? Feb 5 '14 at 17:38

Your solution is correct. Noe that $a=1,b-c=11$ and $a=11, b-c=1$ both lead to $a+b=12+c$. Then $33=ac+bc=(12+c)c$ indeed has no solution. [There's an alomost-soution: $c=-1$ gives $-33$; so I wonder if maybe part of the problem statement may be miscopied]
The only thing I would be a bit more careful about is that if two integers multiply to get $11$, it may be that they are $-1$ and $-11$. For instance, in $a(b-c) = 11$, we could have $b - c = -11$. But as it turns out this cannot work because the other factor, $a$, has to be positive.