{3x+2y = 1, 3x^2+4 = y^2}
Solve 3x+2y = 1 for y
Substitute in
Substitute (-1+2i) in
One solution is (-1+2i, 2-3i)
Substituting (-1-2i) the same way gives 2+3i
So the other solution is (-1-2i, 2+3i)
These solutions are complex imaginary, so the graphs
should not cross:
And, as we see the black line, which is the graph of
, does not cross the red hyperbola, which is
the graph of
Edwin