2006 SMT/Algebra Problems/Problem 6

Problem

Let $a, b, c$ be real numbers satisfying:

$ab-a=b+119$ $bc-b=c+59$ $ca-c=a+71$

Determine all possible values of $a+b+c$ .

Solution

From the first equation, we have $a(b-1)=b+119\implies a=\frac{b+119}{b-1}$ . Plugging this into the third equation, we get $c\left(\frac{b+119}{b-1}\right)-c=\frac{b+119}{b-1}+71$ . Multiplying both sides by $b-1$ , we get $c(b+119)-c(b-1)=b+119+71(b-1)\implies 120c=72b+48\implies c=\frac{3b+2}{5}$ .

Now we plug that into the second equation. We have $b\left(\frac{3b+2}{5}\right)-b=\frac{3b+2}{5}+59$ . Getting rid of the fractions, we have $b(3b+2)-5b=3b+2+295\implies 3b^2-6b-297=0\implies b^2-2b-99=0$ . We can factor that as $(b-11)(b+9)=0$ , so $b=11$ or $b=-9$ .