1. ## Polynomials simultaneous solution

Solve the following simultaneous quadratic equation

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

I have tried solving the equation by substituting $y=mx$ after simplifying both equations above.

I would like to check my answers.... working with latex is troublesome.. it would be great if you could just share the answers out...

$x= +/- \frac{4}{\sqrt 13}$
$y= +/- \frac{3}{\sqrt 13}$

and

$x= +/- \frac{5}{\sqrt 351}$
$y= +/- \frac{4}{3 \sqrt 39 }$

2. Originally Posted by ecogreen
Solve the following simultaneous quadratic equation

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

I have tried solving the equation by substituting $y=mx$ after simplifying both equations above.

I would like to check my answers.... working with latex is troublesome.. it would be great if you could just share the answers out...

$x= +/- \frac{4}{\sqrt 13}$
$y= +/- \frac{3}{\sqrt 13}$

and

$x= +/- \frac{5}{\sqrt 351}$
$y= +/- \frac{4}{3 \sqrt 39 }$
No, that is not correct. Simply substitute $x= \frac{4}{\sqrt 13}$ into both equations and your $y$ will be different both times. You can do the same for your other three $x$-terms and the same thing will happen.

To solve the problem, get rid of the ugly $xy$ term in the middle;

$3(2x^2 +xy-2y^2) + (x^2-3xy-5y^2)=3(-2) + 5$
$\Rightarrow 7x^2-11y^2 = -1$
$\Rightarrow 11y^2 - 7x^2 = 1$.

Now things get interesting. Have you ever heard of the Euclidean algorithm?

(With respect to LaTeX, \pm gives you $\pm$ which is neater than $+/-$.)