# Thread: solving systems of equations

1. ## solving systems of equations

solve the following for u and t

x= u + t^2

y = t - u^2

2. Originally Posted by scubasteve123
solve the following for u and t

x= u + t^2

y = t - u^2
$\displaystyle u=x-t^2$

sub 1st equation into 2nd equation

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

then simplify.

3. 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.

4. 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:

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

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

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

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.

5. 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 :

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