System of equations solution step by step

Question

I'm trying to solve the given system of equations: $$ax+by+(c-1)z=0 $$ $$a-b+2c=0$$ $$x-y+2z=0$$ I need to find the value for the set $\{a,b,c\}$ that holds true for any $\{x,y,z\}$ that follows the $3$rd equation. By trial and error, I was able to find that $\{-1/3, 1/3, 1/3\}$ satisfies the first two equations for all $\{x,y,z\}$ satisfying third equation but I'm not able to arrive at the solution since substituting goes on an endless loop. Can some help with the steps to arrive at the solution? Thanks in advance!

Answer 1

One way to proceed is to solve the last equation for and $z$ and substitute into the first to obtain, after rearrangement, $$\frac12(2a-c+1)x+\frac12(2b+c-1)y=0.$$ This equation must hold for all $x$ and $y$, which can only occur if both coefficients vanish. Together with the second equation, that gives you a system of three linear equations in the unknowns $a$, $b$ and $c$ to solve.