Are there always nontrivial real solutions to $A_1 x^5 + B_1 y^5 + C_1 z^5 = 0$ and $A_2 x + B_2 y + C_2 z = 0$? Firstly, are there always nontrivial real solutions to the sysytem
of equations, $A_{1}x^{5}+B_{1}y^{5}+C_{1}z^{5}=0$ and $A_{2}x+B_{2}y+C_{2}z=0$,
for real numbers $A_{1}$, $B_{1}$, $C_{1}$, $A_{2}$, $B_{2}$,
and $C_{2}$? [Answered]
Secondly, are there always nontrivial real solutions to the sysytem
of equations, $A_{1}\frac{x^{5}}{\left|x\right|}+B_{1}\frac{y^{5}}{\left|y\right|}+C_{1}\frac{z^{5}}{\left|z\right|}=0$
and $A_{2}x+B_{2}y+C_{2}z=0$, for real numbers $A_{1}$, $B_{1}$,
$C_{1}$, $A_{2}$, $B_{2}$, and $C_{2}$? [Answered]
 A: 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 $y$):

A: Yes.  Suppose otherwise and let $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (v_1, v_2, v_3)$ be two linearly independent points on the hyperplane which do not intersect the quintic.  Then some point of the form $\mathbf{u} + \mathbf{v} t$ must intersect the quintic because the corresponding polynomial in $t$ is of degree $5$ (in particular, its leading and constant coefficients are nonzero).  
