I have been unable to solve the above problem.
I will detail my attempt below:
My Solution
$$x^2 + y^2 = 13 \tag{1}$$
$$x^2 - 3xy + 2y^2 = 35 \tag{2}$$
Let $y = mx$.
Rewrite $(1)$ and $(2)$.
$$x^2 + m^2x^2 = 13 \tag{1.1}$$
$$x^2 - 3mx^2 + 2m^2x^2 \tag{2.1}$$
Factorise the above equations.
$$x^2(m^2 + 1) = 13 \tag{1.2}$$
$$x^2(2m^2 - 3m + 1) = 35 \tag{2.2}$$
Divide $(1.2)$ by $(2.2)$ $$\frac{x^2(m^2 + 1)}{x^2(2m^2 - 3m + 1)} = \frac{13}{35}$$ $$\frac{(m^2 + 1)}{(2m^2 - 3m + 1)} = \frac{13}{35}$$
Cross multiply. $$35(m^2 + 1) = 13(2m^2 - 3m + 1)$$ $$35m^2 + 35 = 26m^2 - 39m + 13$$ $$9m^2 + 39m + 22 = 0$$ $$9m^2 + 18m + 11m + 22 = 0$$ $$9m(m + 2) +11(m + 2) = 0$$ $$(9m + 11)(m + 2) = 0$$ $$ m = -2 \text{ or } m = \frac{-11}{9}$$
When $m = -2$ Substitute $m = -2$ into $1.2$ $$x^2(4 + 1) = 13$$ $$5x^2 = 13$$ $$x^2 = \frac{13}{5}$$ $$x = \sqrt{\frac{13}{5}}$$ $$y = -2\sqrt{\frac{13}{5}}$$
When $m = -\frac{11}{9}$ Substitute $m = -\frac{11}{9}$ into $1.2$ $$x^2(\frac{121}{81} + 1) = 13$$ $$x^2(\frac{202}{81}) = 13$$ $$x^2 = \frac{13 * 81}{202} = \frac{1053}{202}$$ $$x = 2.283717322$$ $$y = \frac{-11}{9} * 2.283717322 = -2.790542727$$.
However, when I plug my values for $x$ and $y$ back into $(1)$ and $(2)$, I get $13$ and $39.9$ respectively.
No matter how many times I go over the equation, I just can't solve it satisfactorily, such that when I plug the values back into the equations, I get $13$ and $35$.
What mistakes did I make, and please write out the correct solution.