How do you solve a simultaneous equation with 3 variables that only has two equations? Most of the other requests for solving simultaneous equations seem to have three equations to use, which my specific problem doesn't have.
I also hope to use beginner A-level mathematics, so no calculus, which I've seen in other solutions. The question is:
Find the set of values of k for which the line $y = 2x - k$ meets the curve 
$y = x^2 + kx - 2$ at two distinct points.
If you could explain any solutions you might have, that would be great. 
Edit: thank you all for your the solutions! It has definitely helped. My apologies for saying that calculus was a method of solving, I don’t know the first thing about calculus so thank you for correcting me.
 A: Setting
\begin{eqnarray}
x^2+kx-2&=&2x-k\\
x^2+(k-2)x+(k-2)&=&0\\
\end{eqnarray}
The discriminant must be positive in order to obtain two distinct solutions, so
\begin{eqnarray}
(k-2)^2-4(k-2)&>&0\\
(k-2)(k-6)&>&0
\end{eqnarray}
The graph of this expression with respect to $k$ is a parabola which is concave up, so the expression is positive when $k>6$ and when $k<2$. So the solution set for $k$ is
$$(-\infty,2)\cup(6,\infty)$$
A: Hint: Solving the equation
$$0=x^2+x(k-2)+k-2$$ we get by the quadratic formula
$$x_{1,2}=-\frac{k-2}{2}\pm\sqrt{\left(\frac{k-2}{2}\right)^2-k+2}$$
Simplifying the discriminant
$$\frac{k^2-4k+4-4k+8}{4}=\frac{k^2-8k+12}{4}$$
Can you finish?
A: Well, let's substitute $2x-k$ for $y$ in your quadratic equation, and see what we find;
$$2x-k=x^2+kx-2\to x^2+(k-2)x +(k-2) = 0$$For simplicity, I'll denote $n = k-2$. So, we seek the solutions of the quadratic $$x^2+nx+n=0$$By the quadratic formula, we find that $$x=\frac{-n\pm\sqrt{n^2-4n}}2$$We want to find in which cases there are two solutions to this quadratic, which happens when the discriminant is positive. Hence, we have the inequality $n^2-4n>0$. This implies $$\color{red}{k<2\textrm{ or }k>6}$$
A: The solution range can be determined from the values of k when the line is tangent to the parabola.
$x^2 + kx - 2 = 2x - k\ $  becomes..
$x^2 + (k-2)x -2 + k = 0$
Putting this in perfect square form, $(x + \frac{k-2}{2})^2 = 0$
In which case, $(\frac{k-2}{2})^2 = -2 + k\ $ determines the k values where the line is tangent to the parabola
$(k-2)^2 = -8 + 4k$
$k^2 - 4k + 4 = -8 + 4k$
$k^2 - 8k + 12 = 0$
$(k - 2)(k - 6) = 0$
Therefore $k = 2$ and $k = 6$ are where the line is tangent to the parabola.
Looking at $x^2 + (k-2)x -2 + k = 0$ when $2 < k < 6$, we have all positive terms and hence no roots so the line does not intersect the parabola.
Hence the values of k which intersect the parabola twice are:
$$k<2\ \text{and}\ k>6$$
