
Solving system
http://i146.photobucket.com/albums/r...6/problem6.jpg
I tried to simplify it down and go to this point:
http://i146.photobucket.com/albums/r...problem61.jpg
I don't know if I did it right. From there, I have no idea what to do T_T The problem asks for a exact answer so everything must stay in radicals. I also don't know how to solve it using matrices...

from the first equation ...
$\displaystyle \sqrt{3x} = 2y1$
$\displaystyle 3x = (2y1)^2$
$\displaystyle x = \frac{(2y1)^2}{3}$
sub for x in the second equation ...
$\displaystyle \frac{2(2y1)^2}{3} + 5y^2 = 12$
$\displaystyle 2(2y1)^2 + 15y^2 = 36$
expand, collect like terms, and solve the quadratic for y ... then go back and evaluate for x. note that x > 0.

I solved for y and got
http://i146.photobucket.com/albums/r...6/problem7.jpg
I plugged it into the x equation to solve for x and it looks like this:
http://i146.photobucket.com/albums/r...problem71.jpg
Im not sure how to break that down =O Do I multiply the 2 by both the 4 and root 3192? What about the plus and minus?

why bother? leave those ugly solutions as they are.
use your calculator to check the two sets of solutions (x,y) in the original system.

I’m pretty sure my y is correct. However when I plugged x and y into one of the equations in my calculator, it didn’t work O.o I plugged it in using parentheses in the right places too.
Thanks for the help!