# Thread: Finding x and y

1. ## Finding x and y

Hi,

I have two equations where I need to find x and y but I have no idea how.

$\displaystyle 3e^x + 3xe^x - y^2 + 3y = 0$

and

$\displaystyle e^x(-2y + 3) + 3x - y^2 + 3y = 0$

I am unsure whether logs would be much use or if they would make it worse.

2. $\displaystyle 3e^x + 3xe^x = e^x(-2y + 3) + 3x$

$\displaystyle 3e^x + 3xe^x = 3e^x - 2ye^x + 3x$

$\displaystyle 3xe^x + 2ye^x = 3x$

$\displaystyle e^x(3x + 2y) = 3x$

$\displaystyle e^x = \frac{3x}{3x + 2y}$

Substitute back into the first equation:

$\displaystyle (3x + 3)\left(\frac{3x}{3x + 2y}\right) - y^2 + 3y = 0$

$\displaystyle (3x + 3)(3x) + (3x + 2y)(3y - y^2) = 0$

$\displaystyle 9x^2 + 9x + 9xy + 6y^2 - 3xy^2 - 2y^3 = 0$

Not pretty, but it gets rid of $\displaystyle e^x$ quite nicely. All solutions to this equation such that $\displaystyle \frac{3x}{3x + 2y} > 0$ are the solutions to the original system of equations.

3. I'd imagine the answer to

$\displaystyle What\ happens\ when\ an\ unstoppable\ object\ meets\ an\ immovable\ target$

is

$\displaystyle The\ unstoppable\ object\ doesn't\ stop$

and

$\displaystyle The\ immoveable\ target\ doesn't\ move$

4. Originally Posted by Archie Meade
$\displaystyle What\ happens\ when\ an\ unstoppable\ object\ meets\ an\ immovable\ target$
$\displaystyle The\ unstoppable\ object\ doesn't\ stop$
$\displaystyle The\ immoveable\ target\ doesn't\ move$