Why is eliminating some variables in this system of equations not informative for determining the number of points of intersection?

Question

How many points do $(a) x+2 y=x^2-6$ and (b) $3 x-y=2 x^2+5$ intersect at when both equations are graphed on the $x y$-coordinate plane?

$3x+6y=3x^2-18$

$3x-y=2x^2+5$

$7y=x^2-23 \implies \dfrac{x^2-23}{7}=y$

How do you make any conclusions about the number of points of intersection based on that? If that's useless equation, why? If that's useless, why are other elimination methods more useful?

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$x+2y=x^2-6$

$6x-2y=4x^2+10$

$7x=5x^2+4$

$0=5x^2-7x+4$

The discriminant tells us zero solutions, so no intersections.

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$2x+4y=2x^2-6$

$3x-y=2x^2+5$

$-x+5y=-6-5$

$y=\dfrac{-11+x}{5}$

How do you make any conclusions about the number of points of intersection based on that? If that's useless equation, why? If that's useless, why are other elimination methods more useful?

Answer 1

You need to actually eliminate some variable.

In your first and third method, the last equation still has both $x$ and $y$ - hence no elimination has happened.

In your second method, you have eliminated $y$ and correctly found out that there are no real solutions.

Answer 2

The Bezout theorem tells you that the product of the total degrees of the polynomials is an upper bound for the number of complex solutions. The "theorem" is for the statement that the bound is exact if you also count solutions at infinity.

So in your example systems in the original version you get a bound of $4$. But if you eliminate $x^2$, you can replace one equation with a linear one, giving a bound of $2$ solutions.