How to solve these simultaneous equations using any better way?

Question

This problem is easy, but I am just curious whether there is any better and more elegant method of solving.

Solve the simultaneous equations: $$2x^2+5xy+2y^2=0,$$ $$x^2-y^2=1.$$

The way I solve it is as follows:

$y^2=x^2-1$ (or we can express $x^2$ first). So $2x^2+5xy+2(x^2-1)=0$. Then solve for $y$, then substitute back into $x^2-y^2=1$ to find for $x$. There will be 4 solutions.

Is there any other way to solve this problem? Many thanks!

Answer 1

Hint: $2x^2+5xy+2y^2=(2x+y)(x+2y)$

Answer 2

HINT:

As $x=\sec\theta, y=\tan\theta$ is a parametric form for the second Equation,

express $x,y$ in terms of $\theta$ in the first equation and multiply by $\cos^2\theta$

Answer 3

For the system of equations:

$$\left\{\begin{aligned}&2x^2+5xy+2y^2=0\\&x^2-y^2=1\end{aligned}\right.$$

Has a solution, when $j=\sqrt{-1}$ there is the imaginary unit.

$$x=\frac{\pm2}{\sqrt{3}}$$

$$y=\frac{\mp1}{\sqrt{3}}$$

$$...$$

$$x=\frac{\pm{j}}{\sqrt{3}}$$

$$y=\frac{\mp2j}{\sqrt{3}}$$

Answer 4

As far as you have such relationship in your second equation:

$${x^2} - {y^2} = 1$$

then it comes to mind that one may use the hyperbolic functions to solve this:

$$\begin{array}{l}x = \cosh\left( \theta \right)\\y = \sinh\left( \theta \right)\end{array}$$

by substituting the above formulas to the first equations, you will reduce the two independent variables to just one which is $\theta$;

For ease of calculations and simplifications; you can also use the following relations for hyperbolic functions:

$$\begin{array}{l}x = \cosh\left( \theta \right) = \frac{{{e^\theta } + {e^{ - \theta }}}}{2}\\y = \sinh\left( \theta \right) = \frac{{{e^\theta } - {e^{ - \theta }}}}{2}\end{array}$$

Now, we put it in your first equation;

$$2x^2+5xy+2y^2=0$$ $$2{\left[ {\frac{{{e^\theta } + {e^{ - \theta }}}}{2}} \right]^2} + 5\left[ {\frac{{{e^\theta } + {e^{ - \theta }}}}{2}} \right]\left[ {\frac{{{e^\theta } - {e^{ - \theta }}}}{2}} \right] + 2{\left[ {\frac{{{e^\theta } - {e^{ - \theta }}}}{2}} \right]^2} = 0$$

and we start to simplify it.

$$\begin{array}{l}2\left[ {{e^{2\theta }} + {e^{ - 2\theta }} + 2} \right] + 5\left[ {{e^{2\theta }} - {e^{ - 2\theta }}} \right] + 2\left[ {{e^{2\theta }} + {e^{ - 2\theta }} - 2} \right] = 0;\\\left( {2 + 5 + 2} \right){e^{2\theta }} + \left( {2 - 5 + 2} \right){e^{ - 2\theta }} = 0;\\9{e^{2\theta }} - {e^{ - 2\theta }} = 0;\\9{e^{4\theta }} = 1;\\\theta = \frac{1}{4}\ln\frac{1}{9} = \frac{{ - 1}}{2}\ln3 = \ln\frac{1}{{\sqrt 3 }}.\end{array}$$

so we can easily calculate $x$ and $y$;

$$x = \frac{1}{2}\left( {\frac{1}{{\sqrt 3 }} + \sqrt 3 } \right),y = \frac{1}{2}\left( {\frac{1}{{\sqrt 3 }} - \sqrt 3 } \right);$$

and the equations is solved;

$$x = \frac{2}{{\sqrt 3 }};y = \frac{{ - 1}}{{\sqrt 3 }}.$$