from the 2008 exam 2:
the simultaneous equations 
and 
have a unique solution for:
answer is m: R\{-6,6}
i didnt understand what they were saying in the assessment report :S
If
then clearly
is the unique solution. So assume that
.
Then we multiply the first expression by
and the second by
such that the simultaneous equations read
and
. Subtracting the former from the latter, we find that
y=m^2 - 72)
.
Now if
then this is saying that
, which is impossible. Thus there cannot be a solution if
.
If
, then we can divide through by
to find that
, which we can then substitute back into the original equation to find the corresponding
-coordinate
. This is the unique solution for
.
Graphically, the reason why the simultaneous equations don't have a solution when
is because the lines are parallel at that point: substituting these values of
back into the original equations and making
the subject yields the pairs of equations
and
, which, as they have different
-intercepts, are clearly parallel.
On this thought, you could really just rearrange them to be the equations of lines. If
then clearly
is the unique solution. So assume that
. Then the original equations can be rearranged to read
and
. These have the same gradient when
, which is the same as when
. In this case, however, they have different
-intercepts, and so are clearly parallel, and in Euclidean geometry, parallel lines never intersect, so there is no solution. If, on the other hand,
, then the lines aren't parallel. As straight non-parallel lines intersect exactly once in Euclidean geometry, it follows that there exists a unique solution given by the coordinates of the point of intersection.
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