solve the following for u and t

x= u + t^2

y = t - u^2

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- Nov 10th 2009, 11:10 PMscubasteve123solving systems of equations
solve the following for u and t

x= u + t^2

y = t - u^2 - Nov 10th 2009, 11:17 PMBabyMilo
- Nov 12th 2009, 10:14 PMscubasteve123
I tried this and was just left with y + x^2 = t +2xt^2 - t^4

... not sure how to simplify this I just end up going in circles with all the substitution. - Nov 12th 2009, 11:21 PMBabyMilo
you are right what you are doing.

after you simplfy you get:

$\displaystyle y=t-x^2+2t^2x-t^4$

well we need to rearrange it a little bit, so we get:

$\displaystyle y=-x^2+2t^2x+(t-t^4)$

then using the Quadratic equation

http://upload.wikimedia.org/math/f/4...d3837dfe5a.png

we sub in the numbers,

$\displaystyle \frac{-2t^2\pm\sqrt{(2t^2)^2-4*-1*(t-t^4)}}{2*-1}=0$

then again simplfy:

$\displaystyle \frac{-2t^2\pm\sqrt{4t}}{-2}=0$

so we get t=1 or t=0

if t=0 x=0

if t=1 x=2 or 0.

hope this helps. correct me if im wrong anyone. - Nov 12th 2009, 11:30 PMBacterius
You can simplify even further, Babymilo :

$\displaystyle \frac{-2t^2\pm\sqrt{4t}}{-2}=0$

Extract the $\displaystyle 4$ from the square root :

$\displaystyle \frac{-2t^2\pm 2 \sqrt{t}}{-2}=0$

Factorize $\displaystyle 2$ :

$\displaystyle \frac{2(-t^2 \pm \sqrt{t})}{-2}=0$

Cancel out the $\displaystyle 2$ :

$\displaystyle \frac{-t^2 \pm \sqrt{t}}{-1}=0$

Finally :

$\displaystyle t^2 \mp \sqrt{t} = 0$