# solving systems of equations

• November 11th 2009, 12:10 AM
scubasteve123
solving systems of equations
solve the following for u and t

x= u + t^2

y = t - u^2
• November 11th 2009, 12:17 AM
BabyMilo
Quote:

Originally Posted by scubasteve123
solve the following for u and t

x= u + t^2

y = t - u^2

$u=x-t^2$

sub 1st equation into 2nd equation

$y=t-(x-t^2)^2$

then simplify.
• November 12th 2009, 11:14 PM
scubasteve123
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.
• November 13th 2009, 12:21 AM
BabyMilo
Quote:

Originally Posted by scubasteve123
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.

you are right what you are doing.

after you simplfy you get:

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

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

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

we sub in the numbers,

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

then again simplfy:

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

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.
• November 13th 2009, 12:30 AM
Bacterius
You can simplify even further, Babymilo :

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

Extract the $4$ from the square root :

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

Factorize $2$ :

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

Cancel out the $2$ :

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

Finally :

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