What is wrong with my approach to finding solution of these equations?

Question

I have to find solutions $ \mathbf {(x,y)}$ where $ \mathbf x$ and $ \mathbf y$ are real numbers for the system of equations

$\mathbf {x^2-xy+y^2=21}$

$\mathbf {x^2+2xy-8y^2=0}$

what i initially did was to multiply first equation with 2 and added that to the second equation and that gave me $\mathbf {3x^2-6y^2 = 42}$ this equation for every real $ \mathbf y$ gives me a real $ \mathbf x$ , so i concluded there would be infinite solutions to the above system of equations.

However the answer is that only 4 real solutions exist and i have the solution with me , but i wanted to know why is my method of solving it wrong. Is simple algebraic addition/subtraction of functions to find their intersection not correct? or is it giving me wrong answers only in this specific case?

Answer 1

$x^2 + 2xy - 8 y^2 = 0$ is a pair of lines through the origin. The discriminant is $4 + 32 = 36$ which is a square, so that the form factors $$ x^2 + 2xy - 8 y^2 = (x+4y)(x-2y) $$ and the lines are $x=2y$ and $x=-4y$

$x^2 - xy + y^2 = 21$ is an ellipse. In fact, as $21 = 3 \cdot 7$ we know there are integer points on the ellipse, such as $(4,-1)$ and $(-4,1)$ which are on one of the lines. The other pair of poin ts need extra work and probably square root signs, when $x=2y$ and $x^2 - xy + y^2 = 21$ we are stuck with $4y^2 -2 y^2 + y^2 = 21$ or $3 y^2 = 21,$ or $y = \pm \sqrt 7.$ Thus $(2 \sqrt 7, \sqrt 7)$ and $(-2 \sqrt 7, -\sqrt 7)$

Answer 2

I've graphed all three equations in WolframAlpha for you - the blue and red lines are the given equations, while the yellow is the one you derived.

Notice that all three equations meet at the same four points. What this means is that for the four points which form the solution set of the original equation, the third equation is also true - it's necessary that any point that satisfies both of the given equations also satisfies the derived one - but it's not sufficient, meaning that not every point on $3x^2-6y^2=42$ is guaranteed to be in the solution set.