Is there any solutions for this set of equation? Is there any solutions？ Why?How?
$$
\begin{array}{cccc}
3z &amp;+ &amp;4x + 3y &amp;= 12\\
z &amp;+ &amp;4x + 3y &amp;= 12\\
2z &amp;+ &amp;4x + 3y &amp;= 12\\
\end{array}
$$
Whatever the values of $x$ and $y$, $3z=z=2z$ must hold, and $$z=0.$$ Then the system reduces to

$$4x+3y=12.$$
Yes!! There are many solutions to this set of equations 
3z +4x +3y = 12z +4x +3y 
=> 3z = 12z 
=> z= 0 
Now, since 12 2z + 4x+3y =12 
=> 4x +3y =12  ( as z=0) 
=> 4*0 + 3*4 =12 ie x=0,y=4 ,z=0 
Or, 4*6/4 +3*2 =12 ie, x=3/2, y=2, z=0 
Or, 4*0 + 3*4 = 12 ie, x=0, y= 4, z= 0 
This way, x,y,z can hold infinite sets of values