From the equations \[\begin{align*} 3x + 2y &= 5, \\ 3x^2 + 2y^2 &= k \end{align*}\]

derive an equation relating \(x\) and \(k\).

By rearranging the first equation, we have that \[ y = \frac{5 - 3x}{2}. \] By substituting this into the second equation, we see that \[\begin{align*} 3x^2 + 2 \left( \frac{5 - 3x}{2} \right)^2 = k &\iff 3x^2 + \frac{(5 - 3x)^2}{2} = k \\ &\iff 3x^2 + \frac{25}{2} - 15x + \frac{9}{2} x^2 = k \\ &\iff \frac{15}{2} x^2 - 15x + \frac{25}{2} = k\\ &\iff 15x^2 - 30x + 25 = 2k. \end{align*}\]

Show that there are no real values of \(x\) if \(k < 5\).

Recall that for a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is the expression \(b^2 - 4ac\), and that the quadratic has no real solutions if and only if \(\Delta < 0\).

The previous part of the question led to the quadratic equation \(15x^2-30x+25=2k\), which we can rewrite as \(15x^2-30x+(25-2k)=0\).

The associated discriminant is then \[\begin{align*} (-30)^2 - 4 \times 15(25-2k) &= 900-60(25-2k)\\ &= -600+120k. \end{align*}\]

Therefore there are no real values of \(x\) if and only if \(-600+120k<0\), or \(120k<600\). Dividing by \(120\) gives us \(k<5\) as required.

Geometrically, the two equations represent a fixed straight line and an ellipse of size determined by \(k\). When the ellipse is small enough, the line will not intersect it, as can be seen in this interactivity.