Solve a system of non-linear equations: $2(a-b)=29+4ab$, $2(c-b)= 11+4bc$, $2(c+a) = 9-4ac$ 
So I've solved this system of equations: $$\begin{array}{lcl}2(a-b) & =& 29+4ab \\2(c-b)& = & 11+4bc \\2(c+a) & = & 9-4ac\end{array}$$

by simply solving for each variable in terms of the others, separately and plugging them into the other equations. It does get tedious and I don't want to sit with this for $40$ min in an exam. Is there any efficient way to solve a system like this?

It's $$(2a+1)(2c+1)=10,$$ $$(2b-1)(2c+1)=-12$$ and $$(2a+1)(2b-1)=-30.$$
The rest is smooth:
$$(2a+1)^2(2c+1)^2(2b-1)^2=3600,$$
which gives
$$(2a+1)(2c+1)(2b-1)=60$$ or $$(2a+1)(2c+1)(2b-1)=-60.$$ 
If $(2a+1)(2c+1)(2b-1)=60$ then we obtain: 
$2a+1=\frac{60}{-12}$, $2c+1=\frac{60}{-30}$ and $2b-1=\frac{60}{10}$, which is
$a=-3$, $b=\frac{7}{2}$, $c=-\frac{3}{2}$.
If $(2a+1)(2c+1)(2b-1)=-60$ then the work is similar
and we get the answer:
$$\left\{\left(-3,\frac{7}{2},-\frac{3}{2}\right),\left(2,-\frac{5}{2},\frac{1}{2}\right)\right\}$$
