Without loss of generality, let $A_2\neq0$. Then we have $x=-\frac{B_2}{A_2}y-\frac{C_2}{A_2}z$. Thus we can eliminate $x$ from the first equation to get a degree 5 equation describing a plane curve in $y$ and $z$.

~~By Harnack's curve theorem, the number of components of this curve in the real projective plane is between 1 and 7. Thus you should expect at least one family of solutions to your equations.~~ This paragraph was nonsense because what one really has at this point is an equation relating points on the projective *line*.

**See Karl Schwede's comment or Qiaochu Yuan's answer for a correct characterization**.

Note that their arguments extend to the second question as well. Again, we can look at the $y=1$ slice of the equation we get on eliminating $x$, something like:

$1+C_1\frac{z^5}{|z|}+\frac{(-B_2-C_2z)^5}{|-B_2-C_2z|}=0$

where I've scaled out $A_1,A_2,B_1$. For large positive $z$ and large negative $z$ the function on the left will take opposite signs (which sign is taken will depend on the signs and relative magnitudes of $C_1$ and $C_2$), so you must have at least one root in between.

In the first question, your solutions typically end up as lines through the origin because the homogeneous equation in $y$ and $z$ can be rewritten as one for $y/z$. This doesn't work in your second question and you get much more interesting looking curves (the ordinate is $z$ and the abscissa is $y$):