Consider the following system of equations: $$\begin{cases} 1 + 6(x+y) + 8xy = 0 \\ y = 6x^2 - 8 xy^2 \end{cases} $$
If for example from the first one I solve for $x$, and then I substitute into the second one I get a fourth-degree equation, even a bit painful since it's also complete.
I checked with some program, and I found out that there are four solutions in $\mathbb{C}$, and I mean strictly complex conjugate.
My question is: is there some smart way to figure out in advance, and without programs, if there are no real solutions in this case?
After seeing OP’s comment to the answer posted by @DietrichBurde, we decided to show how to do the same without factorising. Please note that this is not an attempt to belittle the mentioned answer.
The two equation can be written as, $$1+8xy=-6\left(x+y\right)\qquad\text{and}\tag{1}$$ $$y\left(1+8xy\right)=-6x^2.\qquad\qquad\enspace\tag{2}$$
After substituting (1) in (2), we can simplify it to get, $$x^2+y^2+xy=0.\tag{3}$$
After adding $xy$ to both sides of (3), we have, $$\left(x+y\right)^2=xy.\tag{4}$$
We can use (4) to replace $xy$-term in (1). $$8\left(x+y\right)^2+6\left(x+y\right)+1=0\tag{5}$$
From (5), we obtain $$x+y=-\dfrac{1}{4} \quad\text{or}\qquad x+y=-\dfrac{1}{2}.$$
Using these values of $x+y$ in (4) gives, $$xy=\dfrac{1}{16} \quad\text{or}\qquad xy=\dfrac{1}{4}.$$
Assuming $x, y\ne 0$, we can obtain the following pair of equations. $$x+\dfrac{1}{16x}+\dfrac{1}{4}=0\quad\longrightarrow\quad 16x^2+4x+1=0$$ $$x+\dfrac{1}{4x}+\dfrac{1}{2}=0\quad\enspace\longrightarrow\quad\enspace 4x^2+2x+1=0$$
These are the two quadratic factors Dietrich Burde has given in his answer.