Algebra - Solving equations with 3 variables. a,b,c $\in Z^+$ satisfying the equations: $$ 5a + 5b + 2ab = 92$$ $$ 5b+5c+2bc=136$$ $$5c+5a+2ac=244$$ Find the value of 7a + 8b + 9c. 

I took cases like a,b are odd numbers, even numbers and using the properties of their last digits i found the answer. But i want to know if there is some other method to solve this question. 
 A: If you are looking for solutions in integers, try this.
Multiply the first equation by $2$ to obtain $$4ab+10a+10b=184$$ and rewrite as $$(2a+5)(2b+5)=209=11\times 19=-11\times -19= 1\times 209\dots$$
Treat the others the same, and see what drops out.

$(2b+5)(2c+5)=297=11\times 27$ (and various other possible factorisations)  and $(2a+5)(2c+5)=513=27\times 19$
For positive integers $2b+5=11, 2c+5=27, 2a+5=19$
A: In the same sprit as David G. Stork, consider 
$$ 5a + 5b + 2ab = 92\tag 1$$ $$ 5b+5c+2bc=136\tag 2$$ $$5c+5a+2ac=244\tag3$$ 
From $(1)$ elimintate $b$ to get
$$b=\frac{92-5 a}{2 a+5}$$ Plug in $(2)$ and eliminate $c$ to get
$$c=\frac{27 a+201}{19} $$ Plug in $(3)$ and simplify to get
$$\frac{54}{19} (a-7) (a+12)=0$$
A: We can write the system of equations as following $$(2a+5)(2b+5)=209\\(2b+5)(2c+5)=297\\(2a+5)(2c+5)=513$$the first equation has two answers $$a=3,b=7\\a=7,b=3$$the first case is impossible since by substitution is 2nd equation we must have $19|297$ which is impossible therefore $$a=7,b=3$$and we obtain $$c=11$$so $$7a+8b+9c=49+24+99=172$$
A: Two solutions (found by elimination of variables), only one with all positive values:
{a -> 7, b -> 3, c -> 11}
so $7 a + 8 b + 9 c = 172$
A: From the first equation $5a + 5b + 2ab = 92$, we can find:
$$1\le a,b\le 12 \ \ \ \ \ \text{and}\\
a+b=\frac{92-2ab}{5}=18+\frac{2(1-ab)}{5} \Rightarrow ab\equiv 1\pmod{5}.$$
The possible candidates for $(a,b)$ or $(b,a)$ are :
$$(1,6);(1,11);(2,3);(2,8);(3,7);(3,12);(4,9).$$
Only $(3,7)$ or $(7,3)$ satisfy the first equation. 
From the second and third equations one can easily find the complete set.
