have
- an infinite number of solutions,
- no solutions.
Approach 1: using algebra
From the first equation, we can write \[\begin{equation}\label{eq:eq-for-x-in-terms-of-k-and-y} x = \frac{1+ky}{2} \end{equation}\]and substitute this into the second equation, which yields the equation \[ (k+3) \frac{1+ky}{2} - 9y = k. \] After multiplying this through by \(2\) and rearranging, we have that \[ \left( (k+3)k - 18 \right) y + (k+3) - 2k = 0. \] Expanding this leads us to the equation \[ \left( k^2 + 3k - 18 \right) y + 3 - k = 0. \] The quadratic coefficient for \(y\) can be factored, giving the equation \[ (k+6)(k-3)y + 3-k = 0. \]
So \((k+6)(k-3)y-(k-3)=0\), and factorising, we have
\[\begin{equation}\label{eq:linear-equation-in-y-and-k} (k-3)((k+6)y-1)=0. \end{equation}\]Thus \(k = 3\), or \((k+6)y-1=0\).
There are three cases to consider:
Case 1: \(k \ne 3\) and \(k \ne -6\)
In this case, we have a unique solution, namely, \[ y = \frac{1}{k+6} \quad\text{and}\quad x = \frac{(k+6) + k}{2(k+6)} = \frac{k+3}{k+6}. \]
Case 2: \(k = 3\)
If \(k = 3\), equation \(\eqref{eq:linear-equation-in-y-and-k}\) is true for any choice of \(y\), and any choice of \(y\) gives a unique value of \(x\) (namely, that given obtained from equation \(\eqref{eq:eq-for-x-in-terms-of-k-and-y}\)). Thus there are infinitely many solutions \((x,y)\).
Case 3: \(k = -6\)
In this case, equation \(\eqref{eq:linear-equation-in-y-and-k}\) can never hold, as it leads to the contradictory statement that \(9 = 0\). There are therefore no solutions.
Approach 2: Using coordinate geometry
Each of the two equations can be considered as the equation of a straight line.
In general, these two lines will meet in one point.
But if their gradients are equal, then they either have no solutions (if the lines are distinct and parallel) or infinitely many (if the lines coincide).
What are their gradients? The line \(2x - ky = 1\) has gradient \(\dfrac{2}{k}\) (if \(k \neq 0\), and is vertical if \(k=0\)) and \((k+3)x - 9y = k\) has gradient \(\dfrac{k+3}{9}\).
Their gradients are equal if \(\dfrac{2}{k} = \dfrac{k+3}{9}\), where \(k\ne0\), which rearranges to \(k^2 + 3k -18 = 0\). (If \(k=0\), the second line has gradient \(\frac13\), which is not vertical, so the lines are not parallel.)
This factorises to \((k+6)(k-3)=0\), so their gradients are equal when \(k = 3\) and when \(k=-6\).
If \(k = 3,\) then the original equations are \(2x-3y=1\), and \(6x-9y=3\), which coincide, and so there are infinitely many solutions \((x,y)\).
If \(k = -6\), then the original equations become \(2x+6y=1\), or \(x + 3y = 0.5\), and \(-3x-9y=-6\), or \(x+3y =2\). In this case the lines are parallel and distinct, so there are no solutions.
This GeoGebra applet helps us to see what is happening.
