+118687 My algebra was very long-winded and I have not finished it but
I used the first one to get a in terms of b a(b)
then the last one to get b in terms of c b(c)
then i substituted the b(c) into the a to get a(b) in terms of a(c)
Then I substituted this into the middle one and solved for c
etc
There is probably a better way.
+506 Use Simon's favorite factoring trick; it's going to make it so much easier:
\((a+2)(b+2)=-12\\ (a+2)(c+2)=30\\ (b+2)(c+2)=-10\)
Multiply all the equations to get:
\(((a+2)(b+2)(c+2))^2=3600\\ (a+2)(b+2)(c+2)= \pm 60\)
In the first case where the right-hand side of the equation above is positive, to solve for a, divide \((a+2)(b+2)(c+2)=60\) by \((b+2)(c+2)=-10\) to get \(a+2=-6 \rightarrow a=-8\). Do that the same with b and c to get \(b = 0\) and \(c=-7\).
In the second case where the right-hand side of the equation is negative, do the exact same thing mentioned above to get \(a = 4, b = -4, c = 3\).
Therefore, the solutions to this system of the equation are:
\(\boxed{a = -8, b = 0, c = -7 \text{ or } a = 4, b = -4, c = 3} \)
