So it does.
use substitution to solve ...
2x^2 + 3x(x+4) - (x+4)^2 + 1 = 0
2x^2 + 3x^2 + 12x - x^2 - 8x - 16 + 1 = 0
4x^2 + 4x - 15 = 0
(2x-3)(2x+5) = 0
I am sure you can finish it.
THE LINE Y=X+4 INTERSECTS THE CURVE
2X^2 + 3XY - Y^2 + 1 = 0
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