Solve a linear equations with 3 unknowns I have this equations
$$x+3y+6z=3$$
$$x+y+z=-2$$
$$-x+y+4z=7$$
my solution is 
$$x=0$$
$$y=-5$$
$$z=3$$
this task is 1 of 3 and there is going to be 1 that can be solved, 1 that will have a solution with rational numbers and the last 1 wont have any solution. This last one wont have any solution but I have one.
What have I done wrong?

As via Cramer's Rule,
$D=\det\begin{pmatrix} 1 & 3 & 6 \\ 1 & 1 & 1 \\ -1 & 1 & 4\end{pmatrix}=0,$ 
there will be a family of solutions
From the first two equations,
$$x+3y+6z-3=0$$ 
$$x+y+z+2=0$$
Solving for $x,y$ we get $x=\frac{3z-9}2,y=\frac{5-5z}2$
Putting the values of $x,y$ in the third equation, $$-\frac{3z-9}2+\frac{5-5z}2+4z=7\implies 7=7$$
$$\text{So, } x=\frac{3z-9}2,y=\frac{5-5z}2,z=z$$ will satisfy the given three equations for any finite value of $z$
In your case, $z=3$ and for Jerry, $z=5$
