Strategies for linear systems Consider I have the following equations.
Is there a faster way for me to solve the system without going through a series of substitutions? 
$$-20a+13b+13c=0$$
$$10a-26b+13c=0$$
$$10a-13b-16c=0$$
$$a+b+c=1$$

with $$c=1-a-b$$ you will get
$$33a=13$$
$$3a+39b=13$$
$$26a+3b=16$$
this system has no solution.
