A question about Quadratic Simultaneous Equations

Question

$y = 2 - x$

$y = x^2 + 2x + 2$

After substituting $x = 2 - y$ into the second equation (I am aware of substituting y into the second equation), I get $y = 5, y = 2$. However, when I substitute 5 into the second equation ($y = x^2 + 2x + 2$) I get $x = 1, x = -3$ (after solving it using the quadratic formula), where $x=1$ is a wrong answer. However, if I substitute 5 into the first equation ($y = 2 - x$), I get $x=-3$, and after 2 substituted, I get $x = 0$. Why are the answers different when I substitute into the quadratic one? Can't I substitute into the quadratic one?

Answer 1

The blue line $y=2-x$ cuts the parabola $y=x^2+2x+2$ at $(3,-5)$ and $(0,2)$. But when you put y=5 in the second equation, you are actually solving the parabola with the green line $y=5$, which cuts the parabola at 2 points, one of which is $(-3,5)$ and the other is $(1,5)$, which causes your confusion.
This might interest you: Why can a system of quadratic and linear equation produce incorrect solutions?
App used for image: Desmos graphing calculator (just to avoid any consequences of cropping out the logo; I am not affiliated to Desmos in any form whatsoever).

Answer 2

Another way to interpret your method of solution is that inserting $ \ x \ = \ 2 - y \ $ into the quadratic equation leads to $$ y \ \ = \ \ (2 - y)^2 \ + \ 2·(2 - y) \ + \ 2 \ \ \Rightarrow \ \ y \ \ = \ \ y^2 \ - \ 6y \ + \ 10 $$ $$ \Rightarrow \ \ y^2 \ - \ 7y \ + \ 10 \ \ = \ \ (y - 2)·(y - 5) \ \ = \ \ 0 \ \ , $$ which you used to obtain your solutions. This is the equation of a "degenerate" conic, which comprises the two horizontal lines $ \ y \ = \ 2 \ $ and $ \ y \ = \ 5 \ \ . $ (These are seen in the graph in insipidintegrator's answer.)

If you then simply insert the values of $ \ y \ $ into the linear equation, you will have the two intersection points $ \ ( 0 \ , \ 2) \ $ and $ \ ( -3 \ , \ 5) \ $ of the (blue) line with the (red) parabola. But using these values of $ \ y \ $ in the quadratic equation will give you two additional intersections of the "degenerate" conic with the parabola arising from

$ 2 \ = \ x^2 \ + \ 2x \ + \ 2 \ \Rightarrow \ x·(x + 2) \ = \ 0 \ \Rightarrow \ x \ = \ 0 \ , \ -2 \ \ , \ $ producing $ \ (-2 \ , \ 2) \ \ , $ and

$ 5 \ = \ x^2 \ + \ 2x \ + \ 2 \ \Rightarrow \ (x + 3)·(x - 1) \ = \ 0 \ \Rightarrow \ x \ = \ -3 \ , \ 1 \ \ , \ $ for $ \ (1 \ , \ 5) \ \ , $

as you have seen.

There is a theorem (in algebraic geometry) that the number of "finite" intersections of two polynomial curves can be up to the product of the degrees of the polynomials. So there will be up to two (1·2) intersections of a line with a quadratic curve, but as many as four (2·2) intersections between two quadratic curves. We would only be interested in the two intersections on the line, so we "neglect" the other two points you observed.

You can also think of this in terms of the properties of the functions involved. A linear, non-constant function is always one-to-one, so each distinct value of $ \ y \ $ that you found corresponds to a distinct value of $ \ x \ \ $ (two values of $ \ y \ \ , $ two intersection points). But quadratic functions are not one-to-one (except at the "vertex of the parabola"), so a particular value of $ \ y \ $ generally comes from two values of $ \ x \ \ . $ In solving this system then, inserting the values of $ \ y \ $ into the linear function "passes" the "correct" number of intersection points. Inserting them instead back into the quadratic function amounts to solving a different intersection problem.