Combinatorics with 3 equations

Question

Ok, so I want for all combinations of positive integers (a,b,c) for the Equations

$$-10(c-2ab)+\frac{b-5}{a}=52$$ $$a-b+c=53$$ $$a(b+7)=54$$

the product abc

So I want the product abc not the solutions for the euquation aargh, I feel so stupid for not being able to solve this. They are just equations :(

I tried just solving for one variable and then plugging in, but it didn't seem to work :/

Answer 1

Hint: Since $54 = 2\cdot3^3$, we have that $b+7$ may only be $9, 18, 27$ or $54$. Then use the fact that $a$ divides $b-5$.

Answer 2

The equations even over the complex numbers give either $a=3$ or $65a^2 - 89a + 9=0$. Of course, over the integers $a=3, b=11, c=61$. Substituting $b=54/a-7$, we obtain $c=-(a^2 - 46a - 54)/a$ from the second equation, and then $(a-3)(65a^2-89a+9)=0$ form the first equation.