Proof that two simultaneous line equations do not intersect? Apologies if this isn't at the level of questions expected here!
I've got two simultaneous equations to solve.
(Equation 1): $ x y = 4 $
(Equation 2): $ x + y = 2 $
They produce the following curves:

Question: Whilst it's graphically obvious that they do not make contact, what is the algebraic indicator that these two lines do not meet? How do you prove that?
 A: So the curves are
$$\frac4x\quad\text{and}\quad2-x$$
To prove that these curves do not meet, it simply means that
$$\frac4x=2-x$$
has no solutions, to solve this we multiply by $x$
$$4=2x-x^2$$
Then multiply with $-1$ and switch the left and right of the equality
$$x^2-2x=-4$$
Then I'm going to add one to each side
$$x^2-2x+1=-3$$
The left hand side can be rewritten as
$$(x-1)^2=-3$$
No number squared can be negative, so the equation has no solutions.
A: Suppose that the curves meet: there are $x,y$ such that $xy=4$ and $x+y=2$. Then, $x$ and $y$ are real roots to the polynomial
$$
P(z)=z^2-2z+4.
$$
But $P(z)=(z-1)^2+3$ clearly has no real roots, so contradiction. (Ref: Vieta's formulas.)
A: If the two curves intersect (meet), they do so wherever they have a point in common. This will only happen when $y_1=y+2$, \iff  $$\frac 4x = 2-x \iff 2x - x^2 = 4 \iff x^2 -2x + 4 = 0$$ for some real $x$.
Use the quadratic equation to show that there is no real solution to this equation Indeed, you need only check the discriminant of the quadratic $$\underbrace{4 - 16}_{b^2 - 4ac} = -12\lt  0 $$ to see that it is undefined in the reals, and hence no real $x$ that make this equation true. 
I.e., the lines cannot intersect.
