$a, b \in\mathbb{R}$. I have four equations:
$$x+3y-2z+t=-3$$ $$3x+11y+az+5t=2$$ $$3x+12y-6z+6t=b$$ $$4x+15y-8z+8t=-5$$
I have to find out the values of $a$ and $b$ where the system is solvable (has exactly 1 solution).
I also have to find out what values of $a$ and $b$ make the system have infinite solutions and no solutions at all (unsolvable). I know I'm asking for a lot of answers, but this is something where I have absolutely no idea how to solve this, I'd really appreciate the help.
I've made a system of equations which gave me this:
1 Solution:
$$t=-\frac{19}{2},y=15,z=\frac{1}{4}(2x+77),b=\frac{309}{2}, a=-6$$
2 Solution:
$$t=\frac{4}{5}-\frac{b}{15}, x=-\frac{9a(b+38)+58b+1434}{45(a+6)}, y=\frac{1}{45}(4b+57),z=\frac{309-2b}{45(a+6)},a+6\ne0$$